Ray Optics And Optical InstrumentsClass 12 Physics NCERT Solutions

Here, we are given the following information:

1. Position of the Image: The focal length of the mirror is f = R/2 = -36/2 = -18 cm.

Using the mirror equation: 1/v + 1/u = 1/f 1/v + 1/(-27) = 1/(-18) 1/v = 1/(-18) - 1/(-27) = -1/18 + 1/27 1/v = (-3 + 2) / 54 = -1/54 v = -54 cm

Therefore, the screen should be placed 54 cm in front of the mirror to obtain a sharp image.

Now, let us calculate the magnification (m): m = -v/u = -(-54) / (-27) = -2

The size of the image (h') can be calculated as: m = h'/h -2 = h' / 2.5 h' = -2 × 2.5 = -5.0 cm

The size of the image is 5.0 cm, and the negative sign indicates it is inverted.

3. Movement of the Screen: If the candle is moved closer to the mirror (i.e., the magnitude of u decreases from 27 cm), the real image will move farther away from the mirror. For example, when the object is at 2f (-36 cm), the image is also at 2f (-36 cm). As the object moves from -36 cm towards f (-18 cm), the image moves from -36 cm towards infinity. Therefore, the screen would have to be moved away from the mirror.

Here, we are given the following information:

1. Location of the Image: Using the mirror equation: 1/v + 1/u = 1/f 1/v + 1/(-12) = 1/15 1/v = 1/15 + 1/12 1/v = (4 + 5) / 60 = 9/60 = 3/20 v = 20/3 ≈ +6.67 cm

The image is formed at a distance of 6.67 cm behind the mirror. Since v is positive, the image is virtual.

2. Magnification and Nature of the Image: Magnification (m) is given by: m = -v/u = -(20/3) / (-12) = 20/36 = 5/9 m ≈ +0.56

So, the image is virtual, erect, and diminished, located 6.67 cm behind the mirror.

3. Effect of Moving the Needle Farther: As the needle is moved farther from the mirror, the object distance |u| increases. According to the mirror equation for a convex mirror, as u approaches -∞, the image distance v approaches the focal length f (+15 cm).

1/v = 1/f - 1/u As u → -∞, 1/u → 0, so 1/v → 1/f, which means v → f.

Also, the magnification m = f / (f - u). As |u| increases, the denominator (f - u) increases, so the magnification (m) decreases.

Therefore, as the needle is moved farther from the mirror, the image moves from its current position (6.67 cm) towards the focus (15 cm) and becomes progressively smaller in size.

Here, we are given:

Part 1: Refractive index of water (n_w) The refractive index is the ratio of the real depth to the apparent depth.

n_w = Real depth / Apparent depth n_w = 12.5 / 9.4 ≈ 1.33

The refractive index of water is approximately 1.33.

Part 2: Microscope movement for a different liquid Now, the water is replaced by a liquid with the following properties:

First, we calculate the new apparent depth (d'_apparent) of the needle in this liquid: n_l = d'_real / d'_apparent 1.63 = 12.5 / d'_apparent d'_apparent = 12.5 / 1.63 ≈ 7.67 cm

The initial position of the microscope's focus was at the apparent depth in water, which is 9.4 cm from the bottom. The new focus position is at the apparent depth in the liquid, which is 7.67 cm from the bottom.

Since the apparent depth has decreased, the needle appears to be raised further. To focus on the needle again, the microscope must be moved upwards.

Distance the microscope has to be moved = Initial apparent depth - New apparent depth Distance = 9.4 cm - 7.67 cm = 1.73 cm

Therefore, the microscope would have to be moved up by 1.73 cm.

Let's denote the refractive indices of air, water, and glass as n_a, n_w, and n_g respectively. We can take the refractive index of air, n_a = 1.

Using Snell's Law: n_g * sin(i_g) = n_a * sin(r_a) n_g * sin(35°) = 1 * sin(60°) n_g = sin(60°) / sin(35°) = 0.8660 / 0.5736 ≈ 1.51

Using Snell's Law: n_w * sin(i_w) = n_a * sin(r_a) n_w * sin(47°) = 1 * sin(60°) n_w = sin(60°) / sin(47°) = 0.8660 / 0.7314 ≈ 1.18

Using Snell's Law at the water-glass interface: n_w * sin(i) = n_g * sin(r) 1.18 * sin(45°) = 1.51 * sin(r) sin(r) = (1.18 * sin(45°)) / 1.51 sin(r) = (1.18 * 0.7071) / 1.51 sin(r) = 0.8344 / 1.51 ≈ 0.5526 r = arcsin(0.5526) ≈ 33.5°

Alternatively, we can use the relative refractive index without calculating individual indices: Refractive index of glass with respect to water, n_gw = n_g / n_w n_gw = (sin(60°)/sin(35°)) / (sin(60°)/sin(47°)) = sin(47°) / sin(35°)

Now, applying Snell's law for the water-glass interface: n_w * sin(i) = n_g * sin(r) sin(r) = (n_w / n_g) * sin(i) = (1 / n_gw) * sin(i) sin(r) = (sin(35°) / sin(47°)) * sin(45°) sin(r) = (0.5736 / 0.7314) * 0.7071 sin(r) = 0.7843 * 0.7071 ≈ 0.5546 r = arcsin(0.5546) ≈ 33.7°

(The small difference is due to rounding off intermediate values). The predicted angle of refraction in glass is approximately 33.7°.

Here, we are given:

Light from the bulb can emerge out of the water surface only if the angle of incidence at the water-air interface is less than or equal to the critical angle (i_c). If the angle of incidence is greater than the critical angle, total internal reflection occurs, and the light is reflected back into the water.

First, we calculate the critical angle for the water-air interface using the formula: sin(i_c) = 1/n sin(i_c) = 1 / 1.33 ≈ 0.7518 i_c = arcsin(0.7518) ≈ 48.75°

The light from the bulb will emerge from a circular area on the surface of the water. The radius (R) of this circle is determined by the light rays that strike the surface at the critical angle.

From the geometry, we can see a right-angled triangle formed by the depth (d), the radius of the circle (R), and the path of the light ray. In this triangle: tan(i_c) = R / d R = d * tan(i_c)

We know sin(i_c) = 0.7518. We can find cos(i_c) and then tan(i_c). cos(i_c) = sqrt(1 - sin²(i_c)) = sqrt(1 - (0.7518)²) = sqrt(1 - 0.5652) = sqrt(0.4348) ≈ 0.6594 tan(i_c) = sin(i_c) / cos(i_c) = 0.7518 / 0.6594 ≈ 1.14

Now, we can find the radius R: R = 0.8 m * 1.14 = 0.912 m

The area (A) of the surface through which light can emerge is the area of this circle: A = π * R² A = π * (0.912)² A = π * 0.8317 A ≈ 2.61 m²

Therefore, the area of the surface of water through which light from the bulb can emerge is approximately 2.61 m².

Here, we are given:

Part 1: Refractive index of the prism (n_g) The refractive index of the material of the prism (with respect to air, n_a = 1) is given by the formula:

n_g = sin((A + D_m)/2) / sin(A/2)

Substituting the given values: n_g = sin((60° + 40°)/2) / sin(60°/2) n_g = sin(100°/2) / sin(30°) n_g = sin(50°) / sin(30°) n_g = 0.7660 / 0.5 n_g = 1.532

The refractive index of the material of the prism is 1.532.

Part 2: New angle of minimum deviation in water Now, the prism is placed in water.

We need to find the new angle of minimum deviation, let's call it D'_m.

The refractive index to be used in the formula is the refractive index of glass with respect to water (n_gw). n_gw = n_g / n_w = 1.532 / 1.33 ≈ 1.152

Now, we use the prism formula again with this new refractive index: n_gw = sin((A + D'_m)/2) / sin(A/2) 1.152 = sin((60° + D'_m)/2) / sin(60°/2) 1.152 = sin(30° + D'_m/2) / sin(30°) 1.152 = sin(30° + D'_m/2) / 0.5

sin(30° + D'_m/2) = 1.152 * 0.5 = 0.576

Now, we find the angle whose sine is 0.576: 30° + D'_m/2 = arcsin(0.576) 30° + D'_m/2 ≈ 35.17°

D'_m/2 = 35.17° - 30° = 5.17° D'_m = 2 * 5.17° = 10.34°

The new angle of minimum deviation of the parallel beam of light will be approximately 10.34°.

Here, we are given:

According to the Cartesian sign convention for a double-convex lens:

We use the Lens Maker's Formula: 1/f = (n_g/n_a - 1) * (1/R_1 - 1/R_2)

Substituting the given values: 1/20 = (1.55/1 - 1) * (1/R - 1/(-R)) 1/20 = (0.55) * (1/R + 1/R) 1/20 = 0.55 * (2/R) 1/20 = 1.1 / R

R = 1.1 * 20 R = 22 cm

Therefore, the required radius of curvature for both faces is 22 cm.

In this problem, the beam of light is converging at point P. When a lens is placed in its path, point P acts as a virtual object for the lens. The light rays are directed towards P, so the object distance is measured from the lens to P.

Given:

Using the thin lens formula: 1/v - 1/u = 1/f 1/v - 1/12 = 1/20 1/v = 1/20 + 1/12 1/v = (3 + 5) / 60 = 8/60 = 2/15 v = 15/2 = +7.5 cm

Since v is positive, the image is real and is formed 7.5 cm to the right of the lens. The beam will converge at a point 7.5 cm from the lens.

Using the thin lens formula: 1/v - 1/u = 1/f 1/v - 1/12 = 1/(-16) 1/v = -1/16 + 1/12 1/v = (-3 + 4) / 48 = 1/48 v = +48 cm

Since v is positive, the image is real and is formed 48 cm to the right of the lens. The beam will converge at a point 48 cm from the lens.

Here, we are given:

1. Description of the Image: First, let's find the position of the image (v) using the thin lens formula: 1/v - 1/u = 1/f 1/v - 1/(-14) = 1/(-21) 1/v + 1/14 = -1/21 1/v = -1/21 - 1/14 1/v = (-2 - 3) / 42 = -5/42 v = -42/5 = -8.4 cm

Now, let's find the magnification (m) and the size of the image (h'): m = v/u = (-8.4) / (-14) = +0.6

The size of the image is: h' = m * h = 0.6 * 3.0 = 1.8 cm

So, the image is virtual, erect, and diminished, with a size of 1.8 cm, located 8.4 cm in front of the lens.

2. Effect of Moving the Object Further Away: If the object is moved further away from the lens, the object distance |u| increases.

Let's analyze the lens formula: 1/v = 1/f + 1/u. As the object moves further away, u → -∞. Then 1/u → 0. So, 1/v → 1/f, which means v → f.

Also, the magnification m = v/u = f / (f + u). As |u| increases, the denominator (f+u) becomes more negative, and the magnitude of magnification decreases.

Therefore, as the object is moved further away from the lens, the virtual image moves from its current position (-8.4 cm) towards the focal point (-21 cm) and becomes progressively smaller in size.

Here, we are given:

The two thin lenses are in contact. The effective focal length (f) of the combination is given by the formula: 1/f = 1/f_1 + 1/f_2

Substituting the given values: 1/f = 1/30 + 1/(-20) 1/f = 1/30 - 1/20 1/f = (2 - 3) / 60 1/f = -1/60 f = -60 cm

The focal length of the combination is -60 cm.

To determine if the system is converging or diverging, we look at the sign of the effective focal length.

Given:

(a) Final image at the least distance of distinct vision (D) For the eyepiece, the image distance is v_e = -25 cm. Using the lens formula for the eyepiece to find the object distance u_e: 1/v_e - 1/u_e = 1/f_e 1/(-25) - 1/u_e = 1/6.25 -1/u_e = 1/6.25 + 1/25 = (4 + 1)/25 = 5/25 = 1/5 u_e = -5 cm

The image formed by the objective (I_o) serves as the object for the eyepiece. The distance of I_o from the objective is v_o. L_sep = v_o + |u_e| 15 = v_o + 5 v_o = 10 cm

Now, for the objective lens, we find the object distance u_o: 1/v_o - 1/u_o = 1/f_o 1/10 - 1/u_o = 1/2 -1/u_o = 1/2 - 1/10 = (5 - 1)/10 = 4/10 u_o = -10/4 = -2.5 cm The object should be placed 2.5 cm from the objective.

Magnifying power (M): M = m_o * m_e = (v_o/u_o) * (1 + D/f_e) M = (10 / -2.5) * (1 + 25/6.25) M = -4 * (1 + 4) = -4 * 5 = -20 The magnifying power is 20.

(b) Final image at infinity For the final image to be at infinity, the image from the objective (I_o) must be formed at the focal point of the eyepiece. So, the object distance for the eyepiece is u_e = -f_e = -6.25 cm.

The image distance for the objective (v_o) is: L_sep = v_o + |u_e| 15 = v_o + 6.25 v_o = 8.75 cm

Now, for the objective lens, we find the object distance u_o: 1/v_o - 1/u_o = 1/f_o 1/8.75 - 1/u_o = 1/2 -1/u_o = 1/2 - 1/8.75 = (8.75 - 2) / 17.5 = 6.75 / 17.5 u_o = -17.5 / 6.75 ≈ -2.59 cm The object should be placed approximately 2.59 cm from the objective.

Magnifying power (M): M = m_o * m_e = (v_o/u_o) * (D/f_e) M = (8.75 / -2.59) * (25 / 6.25) M ≈ -3.37 * 4 ≈ -13.5 The magnifying power is approximately 13.5.

Given:

The final image is in sharp focus, which implies it is formed at the near point (D = -25 cm).

1. Image distance for the objective (v_o): Using the lens formula for the objective: 1/v_o - 1/u_o = 1/f_o 1/v_o - 1/(-0.9) = 1/0.8 1/v_o = 1/0.8 - 1/0.9 = (0.9 - 0.8) / (0.8 * 0.9) = 0.1 / 0.72 = 1/7.2 v_o = +7.2 cm

2. Object distance for the eyepiece (u_e): The final image is formed at v_e = -25 cm. Using the lens formula for the eyepiece: 1/v_e - 1/u_e = 1/f_e 1/(-25) - 1/u_e = 1/2.5 -1/u_e = 1/2.5 + 1/25 = (10 + 1)/25 = 11/25 u_e = -25/11 ≈ -2.27 cm

3. Separation between the two lenses (L_sep): The separation is the sum of the magnitude of the image distance from the objective and the magnitude of the object distance for the eyepiece. L_sep = |v_o| + |u_e| L_sep = 7.2 + 2.27 = 9.47 cm

The separation between the two lenses is 9.47 cm.

4. Magnifying power of the microscope (M): The magnifying power for an image formed at the near point is given by: M = m_o * m_e = (v_o/u_o) * (1 + D/f_e) M = (7.2 / -0.9) * (1 + 25/2.5) M = -8 * (1 + 10) M = -8 * 11 = -88

The magnifying power of the microscope is 88.

Given:

This is a case of a telescope in normal adjustment, where the final image is formed at infinity. This provides the most comfortable viewing.

1. Magnifying Power (m): The magnifying power of a telescope in normal adjustment is given by the ratio of the focal lengths of the objective and the eyepiece: m = f_o / f_e m = 144 / 6.0 m = 24

The magnifying power of the telescope is 24.

2. Separation between the Objective and the Eyepiece: For a telescope in normal adjustment, the light from a distant object is focused by the objective lens at its focal point. This image then acts as the object for the eyepiece. For the final image to be at infinity, the image from the objective must be located at the focal point of the eyepiece.

Therefore, the separation between the two lenses (the length of the telescope tube) is the sum of their focal lengths. Separation (L) = f_o + f_e L = 144 cm + 6.0 cm L = 150 cm

The separation between the objective and the eyepiece is 150 cm.

Given:

(a) Angular Magnification (m): The angular magnification (or magnifying power) of a telescope in normal adjustment is given by: m = f_o / f_e m = 1500 cm / 1.0 cm m = 1500

The angular magnification of the telescope is 1500.

(b) Diameter of the image of the moon formed by the objective lens: The moon is a very distant object, so the image formed by the objective lens will be at its focal plane, i.e., at a distance v_o = f_o = 15 m from the objective lens.

The angle (α) subtended by the moon at the objective is given by: α ≈ tan(α) = D_moon / u_moon α = (3.48 × 10^6 m) / (3.8 × 10^8 m) ≈ 0.00916 radians

Let the diameter of the image of the moon formed by the objective be d_image. This image is formed at the focal plane of the objective. The angle subtended by the image at the objective lens is also α.

α ≈ tan(α) = d_image / f_o

Therefore, d_image = α * f_o d_image = 0.00916 rad * 15 m d_image ≈ 0.1374 m

d_image = 13.74 cm

The diameter of the image of the moon formed by the objective lens is approximately 13.74 cm.

The mirror equation is: 1/v + 1/u = 1/f Magnification is: m = -v/u

(a) Concave mirror, object between f and 2f: For a concave mirror, f < 0. Let f = -|f|. The object is placed between f and 2f, so 2f < u < f. Since u is negative, -2|f| < u < -|f|. From the mirror equation: 1/v = 1/f - 1/u 1/v = 1/(-|f|) - 1/u Since u is between -2|f| and -|f|, 1/u is between -1/(2|f|) and -1/|f|. So, -1/u is between 1/|f| and 1/(2|f|). Therefore, 1/v = -1/|f| + (a value between 1/(2|f|) and 1/|f|).

(b) Convex mirror, any object location: For a convex mirror, f > 0. The object is real, so u < 0. From the mirror equation: 1/v = 1/f - 1/u Since f > 0 and u < 0, 1/f is positive and -1/u is also positive. Therefore, 1/v is always positive (1/v > 0), which means v is always positive (v > 0). A positive image distance means the image is formed behind the mirror, so it is always virtual.

(c) Convex mirror, image location and size: From part (b), 1/v = 1/f - 1/u. Since u < 0, -1/u > 0. Therefore, 1/v = 1/f + |-1/u| > 1/f. If 1/v > 1/f, then v < f. Since v is also positive, we have 0 < v < f. This means the image is always located between the pole and the focus. For magnification, m = -v/u. Since v is positive and u is negative, m is always positive, meaning the image is erect. m = -v/u. We can write this as m = f / (f - u). Since f > 0 and u < 0, the denominator (f - u) is always greater than f. So, m = f / (f - u) < 1. Since m is also positive, 0 < m < 1. This means the image is always diminished in size.

(d) Concave mirror, object between pole and focus: For a concave mirror, f < 0. Let f = -|f|. The object is between the pole (0) and focus (f), so f < u < 0. This means -|f| < u < 0. From the mirror equation: 1/v = 1/f - 1/u = -1/|f| - 1/u. Since u is negative and -|f| < u, we have 1/u < -1/|f|. Therefore, -1/u > 1/|f|. So, 1/v = -1/|f| - 1/u = (-1/|f|) + (a value greater than 1/|f|). This means 1/v is positive, so v > 0. A positive image distance means the image is formed behind the mirror, hence it is virtual. For magnification, m = -v/u. Since v > 0 and u < 0, the magnification m is positive, meaning the image is erect. Also, since 1/v = 1/f - 1/u, we can write v = uf / (u-f). Then m = -v/u = -f / (u-f). For a concave mirror, f is negative. Let f = -|f|. u is also negative. Let u = -|u|. m = -(-|f|) / (-|u| - (-|f|)) = |f| / (|f| - |u|). Since the object is between the pole and focus, |u| < |f|. Therefore, (|f| - |u|) is positive and less than |f|. This makes m = |f| / (|f| - |u|) > 1. The image is enlarged.

Given:

When viewed through the glass slab, the pin will appear to be at a shallower depth. This is the apparent depth. The formula for apparent depth (d_app) is: Apparent depth = Real depth / Refractive index d_app = t / n d_app = 15 cm / 1.5 = 10 cm

The apparent position of the pin is 10 cm below the top surface of the slab.

The amount by which the pin appears to be raised is called the normal shift (s). Shift (s) = Real depth - Apparent depth s = 15 cm - 10 cm = 5 cm

Alternatively, the formula for normal shift is: s = t * (1 - 1/n) s = 15 * (1 - 1/1.5) = 15 * (1 - 2/3) = 15 * (1/3) = 5 cm

So, the pin would appear to be raised by a distance of 5 cm.

Does the answer depend on the location of the slab? No, the answer does not depend on the location of the slab, as long as the slab is placed between the eye and the pin. The normal shift (s = t * (1 - 1/n)) depends only on the thickness (t) and the refractive index (n) of the slab, not on its position relative to the object or the observer. The apparent position of the pin will change relative to the observer, but the amount it is raised relative to its actual position remains the same.

Given:

(a) With outer covering: For total internal reflection (TIR) to occur at the core-cladding interface, the angle of incidence at this interface (let's call it i') must be greater than the critical angle (i_c).

First, calculate the critical angle i_c: sin(i_c) = n_2 / n_1 sin(i_c) = 1.44 / 1.68 = 6/7 ≈ 0.8571 i_c = arcsin(0.8571) ≈ 59°

So, for TIR, we must have i' > 59°.

Now, consider a ray entering the pipe from air (n_0 = 1) at an angle of incidence i with the axis. Let the angle of refraction inside the core be r. From the geometry shown in the figure, we have r + i' = 90°, so r = 90° - i'.

Apply Snell's law at the air-core interface: n_0 * sin(i) = n_1 * sin(r) 1 * sin(i) = n_1 * sin(90° - i') sin(i) = n_1 * cos(i')

Since we need i' > i_c, this means cos(i') < cos(i_c). Because cos(θ) is a decreasing function for 0 < θ < 90°. Therefore, sin(i) < n_1 * cos(i_c).

We can find cos(i_c) from sin(i_c): cos(i_c) = sqrt(1 - sin²(i_c)) = sqrt(1 - (1.44/1.68)²) cos(i_c) = sqrt(1 - (0.8571)²) = sqrt(1 - 0.7346) = sqrt(0.2654) ≈ 0.515

Now, substitute this into the inequality for sin(i): sin(i) < 1.68 * 0.515 sin(i) < 0.8652

i < arcsin(0.8652) ≈ 59.9°

The maximum angle of incidence is i_max ≈ 59.9°. The minimum angle is 0. So, the range of incident angles is 0 ≤ i < 59.9°.

(b) Without outer covering: If there is no outer covering, the light pipe is in air. So, the cladding medium is air, with n_2 = 1.

First, calculate the new critical angle i_c: sin(i_c) = n_2 / n_1 = 1 / 1.68 ≈ 0.5952 i_c = arcsin(0.5952) ≈ 36.5°

For TIR, we must have i' > 36.5°.

Using the same relation as before: sin(i) = n_1 * cos(i'). We need i' > i_c, which implies cos(i') < cos(i_c). So, sin(i) < n_1 * cos(i_c).

Calculate the new cos(i_c): cos(i_c) = sqrt(1 - sin²(i_c)) = sqrt(1 - (1/1.68)²) cos(i_c) = sqrt(1 - 0.3543) = sqrt(0.6457) ≈ 0.8035

Now, substitute this value: sin(i) < 1.68 * 0.8035 sin(i) < 1.35

Since the maximum value of sin(i) is 1, this inequality is true for all possible angles of incidence (from 0° to 90°). Therefore, any ray entering the pipe from the air will undergo total internal reflection, regardless of the incident angle. The range of angles is 0 ≤ i ≤ 90°.

Let the distance between the object (bulb) and the image (screen on the opposite wall) be D. Here, D = 3 m. Let the distance of the object from the lens be u, and the distance of the image from the lens be v.

From the setup, we can write: Object distance = -x (let's use magnitude x) Image distance = +(D - x) So, u = -x and v = D - x.

Using the lens formula: 1/v - 1/u = 1/f 1/(D-x) - 1/(-x) = 1/f 1/(D-x) + 1/x = 1/f (x + D - x) / (x(D-x)) = 1/f D / (Dx - x²) = 1/f f = (Dx - x²) / D

For a real image to be formed by a convex lens, the lens formula must have a real solution for x. The equation can be rewritten as: x² - Dx + Df = 0

For x to be real, the discriminant of this quadratic equation must be greater than or equal to zero. Discriminant = (-D)² - 4(1)(Df) ≥ 0 D² - 4Df ≥ 0 D² ≥ 4Df D ≥ 4f (since D > 0)

This gives us the condition for the focal length: f ≤ D/4

This means the maximum possible focal length (f_max) of the lens is D/4.

Given D = 3 m: f_max = 3 m / 4 = 0.75 m

Therefore, the maximum possible focal length of the convex lens required for this purpose is 0.75 m.

This is a classic problem related to the displacement method for finding the focal length of a convex lens.

Let:

In this method, the object and screen are fixed. The lens is moved between them. There are two positions of the lens for which a sharp, real image is formed on the screen. Let these positions be L1 and L2.

If the object distance for the first position (L1) is u, then the image distance is v = D - u. For the second position (L2), due to the principle of reversibility of light, the object and image distances are interchanged. So, the new object distance is v, and the new image distance is u.

The distance between the two lens positions is the difference between the object distances in the two cases: d = v - u

We have a system of two equations:

Adding the two equations: 2v = 110 => v = 55 cm

Subtracting the second equation from the first: 2u = 70 => u = 35 cm

Now we have the object distance u = 35 cm and image distance v = 55 cm for one of the lens positions. Using the sign convention, u = -35 cm and v = +55 cm.

We can find the focal length (f) using the lens formula: 1/f = 1/v - 1/u 1/f = 1/55 - 1/(-35) 1/f = 1/55 + 1/35 1/f = (7 + 11) / 385 1/f = 18 / 385 f = 385 / 18 ≈ 21.39 cm

Alternatively, there is a direct formula for the displacement method: f = (D² - d²) / 4D f = (90² - 20²) / (4 * 90) f = (8100 - 400) / 360 f = 7700 / 360 f = 770 / 36 ≈ 21.39 cm

Therefore, the focal length of the lens is approximately 21.4 cm.

From Exercise 9.10, we have:

(a) Effective Focal Length: The formula for the effective focal length (F) of two coaxial lenses separated by a distance d is: 1/F = 1/f_1 + 1/f_2 - d/(f_1 * f_2)

1/F = 1/30 + 1/(-20) - 8.0 / (30 * -20) 1/F = 1/30 - 1/20 + 8.0 / 600 1/F = (20 - 30 + 8) / 600 1/F = -2 / 600 = -1 / 300 F = -300 cm

The effective focal length of the combination is -300 cm.

Dependence on the side of incidence: Yes, for a system of thick lenses or separated thin lenses, the position of the principal planes depends on the direction of incident light. The effective focal length is measured from the second principal plane. If the light comes from the other side, the roles of the lenses are reversed, and the principal planes will be at different locations. However, the value of the effective focal length itself (as calculated by the formula above) does not change. So, the numerical value of F is the same, but its physical meaning (measured from which point) changes.

Usefulness of effective focal length: The notion is useful because it allows a complex system of lenses to be treated as a single equivalent thin lens placed at a specific location (the principal plane). This simplifies calculations for image formation, especially in designing optical instruments.

Image formed by the first lens (convex): 1/v_1 - 1/u_1 = 1/f_1 1/v_1 - 1/(-40) = 1/30 1/v_1 = 1/30 - 1/40 = (4 - 3)/120 = 1/120 v_1 = +120 cm

This image is formed 120 cm to the right of the convex lens. Magnification by the first lens, m_1 = v_1 / u_1 = 120 / (-40) = -3.

Image formed by the second lens (concave): The image from the first lens acts as the object for the second lens. Since the lenses are 8 cm apart, this object is located (120 - 8) = 112 cm to the right of the concave lens. This is a virtual object. Object distance for the second lens, u_2 = +112 cm.

1/v_2 - 1/u_2 = 1/f_2 1/v_2 - 1/112 = 1/(-20) 1/v_2 = -1/20 + 1/112 = (-112 + 20) / (20 * 112) = -92 / 2240 v_2 = -2240 / 92 ≈ -24.35 cm

The final image is formed 24.35 cm to the left of the concave lens. Magnification by the second lens, m_2 = v_2 / u_2 = -24.35 / 112 ≈ -0.217

Total Magnification and Final Image Size: Total magnification, M = m_1 * m_2 M = (-3) * (-0.217) ≈ +0.65

The total magnification produced by the two-lens system is approximately 0.65.

Size of the final image, h_2 = M * h_1 h_2 = 0.65 * 1.5 cm ≈ 0.98 cm

The size of the final image is approximately 0.98 cm. Since the magnification is positive, the final image is erect.

Given:

The ray of light just suffers total internal reflection (TIR) at the second face. This means the angle of incidence at the second face (r_2) is equal to the critical angle (i_c).

1. Calculate the critical angle (i_c): sin(i_c) = 1/n (since the prism is in air) sin(i_c) = 1 / 1.524 ≈ 0.6562 i_c = arcsin(0.6562) ≈ 41°

So, for TIR to just occur, r_2 = i_c = 41°.

2. Find the angle of refraction at the first face (r_1): For a prism, the angle of the prism is related to the angles of refraction at the two faces by the formula: A = r_1 + r_2 60° = r_1 + 41° r_1 = 60° - 41° = 19°

3. Find the angle of incidence on the first face (i): Now, we apply Snell's law at the first face, where the light enters the prism from the air (n_air = 1).

n_air * sin(i) = n * sin(r_1) 1 * sin(i) = 1.524 * sin(19°)

Using sin(19°) ≈ 0.3256: sin(i) = 1.524 * 0.3256 sin(i) ≈ 0.4962 i = arcsin(0.4962) ≈ 29.75°

Therefore, the ray of light should be incident at an angle of approximately 29.75° on the face of the prism.

Given:

(a) Magnification and Image Area: First, we find the position of the image (v) using the lens formula: 1/v - 1/u = 1/f 1/v - 1/(-9) = 1/9 1/v + 1/9 = 1/9 1/v = 0 v = ∞

The image is formed at infinity.

The linear magnification (m) is given by m = v/u. Since v is infinity, the linear magnification is infinite. However, the question might be interpreted differently. Let's assume the lens has a focal length of 10 cm as given in some versions of the textbook, as f=9cm and u=9cm leads to an image at infinity which doesn't allow for a finite magnification calculation. Let's proceed with f=10cm as a common variant of this problem for a more illustrative answer.

Recalculating with f = 10 cm: 1/v - 1/(-9) = 1/10 1/v = 1/10 - 1/9 = (9-10)/90 = -1/90 v = -90 cm

Linear magnification, m = v/u = -90 / -9 = 10.

Area of each square in the virtual image (A_i) is related to the object area by the square of the linear magnification (areal magnification = m²). A_i = m² * A_o A_i = (10)² * 1 mm² = 100 mm²

(b) Angular Magnification (Magnifying Power): Angular magnification (M) is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye when placed at the near point (D = 25 cm). For an image formed at a distance v, the magnifying power is M = D/u (when the eye is close to the lens) for the image at v=-90cm. Or, more generally, M = 1 + D/f for image at near point, and M = D/f for image at infinity. Let's calculate for the image at v=-90cm. The formula is M = D(1/u - 1/v) = 25(1/9 - 1/90) = 25(9/90) = 2.5. Let's use the standard formula for an image at distance v: M = 1 - v/f = 1 - (-90/10) = 1+9=10. This is not the angular magnification. The angular magnification is (angle by image)/(angle by object at D). Angle by image = h'/|v| = hm/|v| = h10/90 = h/9. Angle by object at D = h/D = h/25. Angular magnification = (h/9)/(h/25) = 25/9 = 2.78. Let's stick to the book's definition. The magnifying power when the image is not at infinity or D is less clear. Let's re-evaluate the question with the original data f=9cm, u=9cm.

Re-evaluating with original data (f=9cm, u=9cm): (a) As calculated, v = ∞. The linear magnification m is infinite. The area of the image is also infinite.

(b) The angular magnification (magnifying power) for an image formed at infinity is given by: M = D / f M = 25 cm / 9 cm ≈ 2.78

(c) Comparison of Magnification and Magnifying Power: No, the magnification in (a) is not equal to the magnifying power in (b).

The two concepts are different. Linear magnification relates the actual sizes, while angular magnification relates the apparent angular sizes.

From Exercise 9.22, we have:

(a) Distance of the lens from the card sheet (u): Maximum possible magnifying power is achieved when the final virtual image is formed at the near point of the eye, which is D = 25 cm. So, v = -25 cm.

We need to find the object distance (u) using the lens formula: 1/v - 1/u = 1/f 1/(-25) - 1/u = 1/9 -1/u = 1/9 + 1/25 -1/u = (25 + 9) / 225 = 34 / 225 u = -225 / 34 ≈ -6.62 cm

The lens should be held at a distance of approximately 6.62 cm from the card sheet.

(b) Magnification (m): The linear magnification (m) is given by: m = v/u = (-25) / (-225/34) = 25 * (34 / 225) = 34 / 9 ≈ 3.78

(c) Magnifying Power (M) and Comparison: The magnifying power (M) when the image is formed at the near point is given by: M = 1 + D/f M = 1 + 25/9 = (9+25)/9 = 34/9 ≈ 3.78

Yes, in this specific case, the magnitude of the linear magnification is equal to the angular magnification (magnifying power).

In this case, the image is formed at the near point, so |v| = D. Therefore, M = m * (D/D) = m. So, when the final image is formed at the near point of distinct vision, the linear magnification is numerically equal to the magnifying power.

From previous exercises, we have:

1. Find the required linear magnification (m): The areal magnification is the square of the linear magnification. Areal Magnification = A_i / A_o = m² m² = 6.25 mm² / 1 mm² = 6.25 m = sqrt(6.25) = 2.5

Since the image formed by a magnifying glass is virtual and erect, the magnification is positive, so m = +2.5.

2. Find the object distance (u): We know that m = v/u. So, v = 2.5u.

Now, substitute v in the lens formula: 1/v - 1/u = 1/f 1/(2.5u) - 1/u = 1/9 (1 - 2.5) / (2.5u) = 1/9 -1.5 / (2.5u) = 1/9 -1.5 * 9 = 2.5u -13.5 = 2.5u u = -13.5 / 2.5 = -5.4 cm

The distance between the object (card sheet) and the magnifying glass should be 5.4 cm.

3. Can the image be seen distinctly? Let's find the position of the image (v) corresponding to this object distance: v = 2.5u = 2.5 * (-5.4 cm) = -13.5 cm

The virtual image is formed at a distance of 13.5 cm from the lens. The normal near point for an eye is 25 cm. Since the image is formed at 13.5 cm, which is much closer than the normal near point, a person with normal vision would not be able to see the squares distinctly. The eye would be under a lot of strain and the image would appear blurred.

(a) Sense of Angular Magnification: The statement is correct that the angle subtended by the virtual image at the eye (or lens) is the same as the angle subtended by the object at the eye (or lens). However, the magnifying glass allows us to bring the object much closer to the eye than the normal near point (D ≈ 25 cm). Without the lens, to see the object clearly, we must place it at or beyond D, where it subtends a small angle. With the lens, we can place the object very close (at or within the focal length), where it subtends a much larger angle at the eye. The lens then forms a virtual image at a comfortable viewing distance (like D or infinity). The angular magnification is the ratio of the angle subtended by the image (which is the same as the large angle subtended by the nearby object) to the angle subtended by the object if it were placed at the near point D. Thus, magnification is achieved by enabling a larger viewing angle.

(b) Effect of Moving the Eye Back: Yes, the angular magnification changes if the eye is moved back from the lens. The formulas M = D/f (image at infinity) and M = 1 + D/f (image at near point) are derived assuming the eye is very close to the lens. If the eye is moved back, the angle subtended by the image decreases, while the reference angle (object at D) remains the same. This reduces the angular magnification. For best results, the eye should be positioned as close to the lens as possible.

(c) Limitation on Focal Length: There are practical limitations that stop us from using lenses with extremely small focal lengths:

(d) Short Focal Lengths in Compound Microscope: The total magnification of a compound microscope is approximately M = (L/f_o) * (D/f_e), where L is the tube length, f_o is the objective's focal length, and f_e is the eyepiece's focal length. To achieve a large overall magnification (M), both f_o and f_e must be small. A small f_o produces a highly magnified intermediate image, and a small f_e provides high angular magnification of this intermediate image.

(e) Position of the Eye (Eye-Ring): The best position for viewing is at a specific point called the 'eye-ring' or exit pupil. This is the point where all the rays emerging from the eyepiece, coming from the full field of view of the objective, cross the axis. If the eye is placed at the eye-ring, it can collect the maximum amount of light from the instrument and see the brightest and widest field of view. If the eye is too close or too far from this point, some of the rays will miss the pupil of the eye, resulting in a restricted field of view (vignetting). The eye-ring is essentially the image of the objective lens formed by the eyepiece. Its distance from the eyepiece can be calculated using the lens formula with u being the separation between the lenses and f being the eyepiece focal length.

Given:

The total magnification of a compound microscope is the product of the magnification of the objective (m_o) and the eyepiece (m_e). M = m_o * m_e

First, let's find the magnification of the eyepiece (m_e). For comfortable viewing, we assume the final image is formed at infinity. In this case: m_e = D / f_e = 25 cm / 5 cm = 5

Now, we can find the required magnification from the objective (m_o): M = m_o * m_e 30 = m_o * 5 m_o = 30 / 5 = 6

The objective must provide a linear magnification of -6 (the image is real and inverted).

We know that m_o = v_o / u_o. So, v_o / u_o = -6, which means v_o = -6u_o.

Now, we use the lens formula for the objective to find the object distance (u_o) and image distance (v_o): 1/v_o - 1/u_o = 1/f_o 1/(-6u_o) - 1/u_o = 1/1.25 (-1 - 6) / (6u_o) = 1/1.25 -7 / (6u_o) = 1/1.25 6u_o = -7 * 1.25 = -8.75 u_o = -8.75 / 6 ≈ -1.46 cm

So, the object must be placed approximately 1.46 cm in front of the objective.

Now, let's find the image distance for the objective: v_o = -6u_o = -6 * (-1.46 cm) ≈ 8.75 cm

Setting up the microscope: To get the final image at infinity, the image formed by the objective (at v_o) must be at the focal point of the eyepiece. Therefore, the separation between the objective and the eyepiece (the tube length) should be: Separation (L) = v_o + f_e L = 8.75 cm + 5 cm = 13.75 cm

Given:

(a) Telescope in normal adjustment (image at infinity): In normal adjustment, the magnifying power (m) of the telescope is given by the ratio of the focal lengths of the objective and the eyepiece.

m = f_o / f_e m = 140 cm / 5.0 cm m = 28

The magnifying power in normal adjustment is 28.

(b) Final image at the least distance of distinct vision: When the final image is formed at the near point (D), the formula for the magnifying power (m) is:

m = (f_o / f_e) * (1 + f_e / D)

Substituting the given values: m = (140 / 5.0) * (1 + 5.0 / 25) m = 28 * (1 + 0.2) m = 28 * 1.2 m = 33.6

The magnifying power when the final image is formed at the least distance of distinct vision is 33.6.

From Exercise 9.27:

(a) Separation between the lenses in normal adjustment: In normal adjustment (final image at infinity), the image formed by the objective is at its focal point, which must also be the focal point of the eyepiece. Therefore, the separation between the lenses is the sum of their focal lengths.

Separation (L) = f_o + f_e L = 140 cm + 5.0 cm = 145 cm

For a distant object, the objective lens forms an image at its focal plane. So, the image distance is v_o ≈ f_o = 140 cm = 1.4 m.

The magnification produced by the objective is m_o = v_o / u_o. The height of the image (h_i) is given by h_i = m_o * h_o.

h_i = (v_o / u_o) * h_o h_i = (1.4 m / 3000 m) * 100 m h_i ≈ 0.0467 m = 4.67 cm

Alternatively, the angle subtended by the tower at the objective is α = h_o / u_o. The angle subtended by the image at the objective is α = h_i / f_o. So, h_i / f_o = h_o / u_o => h_i = f_o * (h_o / u_o) = 1.4 * (100/3000) = 1.4/30 ≈ 0.0467 m = 4.67 cm.

The height of the image of the tower formed by the objective lens is 4.67 cm.

(c) Height of the final image (formed at 25 cm): The image formed by the objective (h_i = 4.67 cm) acts as the object for the eyepiece. The eyepiece acts as a simple microscope. The final image is formed at v_e = -25 cm.

The magnification produced by the eyepiece (m_e) is: m_e = 1 + D/f_e = 1 + 25/5 = 1 + 5 = 6

The height of the final image (h_f) is the height of the object for the eyepiece (h_i) multiplied by the eyepiece magnification (m_e).

h_f = m_e * h_i h_f = 6 * 4.67 cm ≈ 28.02 cm

The height of the final image of the tower is approximately 28 cm.

Given:

Step 1: Image formed by the primary mirror (objective) For an object at infinity, the primary concave mirror will form a real image at its focal point. Focal length of the primary mirror, f_1 = R_1 / 2 = -220 / 2 = -110 mm. The image (I_1) would be formed 110 mm in front of the primary mirror.

Step 2: Image formation by the secondary mirror This image I_1 acts as a virtual object for the secondary convex mirror. The secondary mirror is placed 20 mm in front of the primary mirror. The rays are converging towards I_1. The distance of this virtual object from the secondary mirror is: u_2 = (110 - 20) mm = 90 mm. Since it is a virtual object (behind the mirror), its distance is taken as positive according to the sign convention: u_2 = +90 mm.

Focal length of the secondary mirror, f_2 = R_2 / 2 = +140 / 2 = +70 mm.

Step 3: Find the position of the final image (v_2) Using the mirror equation for the secondary mirror: 1/v_2 + 1/u_2 = 1/f_2 1/v_2 + 1/90 = 1/70 1/v_2 = 1/70 - 1/90 1/v_2 = (9 - 7) / 630 = 2 / 630 = 1 / 315 v_2 = +315 mm

The positive sign indicates that the final image is formed to the right of the secondary mirror (real image). The question asks for the location of the final image, which is usually measured from the primary mirror's pole (where there is a hole).

The distance of the final image from the secondary mirror is 315 mm. The distance of the final image from the primary mirror is (315 mm - 20 mm) = 295 mm behind the primary mirror.

Therefore, the final image will be formed at a distance of 315 mm from the small (secondary) mirror.

Given:

According to the laws of reflection, when a plane mirror is rotated by an angle θ, the reflected ray rotates by an angle of 2θ.

So, the angle of deflection of the reflected ray is: Φ = 2θ Φ = 2 * 3.5° = 7.0°

Now, we have a right-angled triangle formed by the mirror, the screen, and the path of the light spot. The angle of deflection is Φ, the adjacent side is the distance to the screen D, and the opposite side is the displacement of the spot on the screen (let's call it x).

We can use the tangent function: tan(Φ) = x / D x = D * tan(Φ) x = 1.5 m * tan(7.0°)

Using the value tan(7.0°) ≈ 0.1228: x = 1.5 * 0.1228 x ≈ 0.1842 m

x = 18.42 cm

Therefore, the displacement of the reflected spot of light on the screen is approximately 18.4 cm.

This experiment uses the principle that if an object is placed at the focal point of a lens system placed on a plane mirror, the rays after passing through the lens become parallel, strike the mirror normally, retrace their path, and form an image at the object's position.

The system consists of two lenses in contact: the glass convex lens (f_lens = 30 cm) and the liquid plano-concave lens (f_liquid). The effective focal length F is given by: 1/F = 1/f_lens + 1/f_liquid

We can find the focal length of the liquid lens: 1/f_liquid = 1/F - 1/f_lens 1/f_liquid = 1/45 - 1/30 1/f_liquid = (2 - 3) / 90 = -1/90 f_liquid = -90 cm

The negative sign confirms that the liquid lens is diverging (plano-concave).

Finding the refractive index of the liquid (n_l): We need to use the Lens Maker's formula for both lenses. For the equiconvex glass lens (n_g = 1.50): Let the radius of curvature be R. R_1 = +R, R_2 = -R. 1/f_lens = (n_g - 1) * (1/R_1 - 1/R_2) 1/30 = (1.50 - 1) * (1/R - 1/(-R)) 1/30 = 0.5 * (2/R) = 1/R R = 30 cm

Now, for the plano-concave liquid lens:

Using the Lens Maker's formula for the liquid lens: 1/f_liquid = (n_l - 1) * (1/R_1 - 1/R_2) -1/90 = (n_l - 1) * (1/(-30) - 1/∞) -1/90 = (n_l - 1) * (-1/30) 1/90 = (n_l - 1) / 30 n_l - 1 = 30 / 90 = 1/3 n_l = 1 + 1/3 = 4/3 n_l ≈ 1.33

The refractive index of the liquid is approximately 1.33.