Light – Reflection and RefractionClass 10 Physics NCERT Solutions

Answer: (d) Clay

Reason: A lens is a transparent optical medium that refracts light. Materials like water, glass, and plastic are transparent and can be shaped to form lenses. Clay, however, is an opaque material, which means it does not allow light to pass through it. Therefore, clay cannot be used to make a lens.

Answer: (d) Between the pole of the mirror and its principal focus.

Reason: A concave mirror forms a virtual, erect, and magnified image only when the object is placed between its pole (P) and principal focus (F). For all other positions of the object in front of the mirror, it forms a real and inverted image (except when the object is at the focus, where the image is formed at infinity).

Answer: (b) At twice the focal length

Reason: For a convex lens, when an object is placed at the center of curvature, which is at a distance of twice the focal length ($2F_1$) from the optical center, it forms a real, inverted image of the same size at twice the focal length on the other side of the lens ($2F_2$).

Answer: (a) both concave.

Reason: According to the New Cartesian Sign Convention, the focal length of a concave mirror is negative, and the focal length of a concave lens is also negative. A convex mirror and a convex lens both have positive focal lengths. Since both the mirror and the lens have a focal length of $-15 \text{ cm}$, they must both be concave.

Answer: (d) either plane or convex.

Answer: (c) A convex lens of focal length 5 cm.

Reason: To read small letters, a magnifying glass is used, which is a convex lens. A convex lens produces a magnified, virtual, and erect image when the object (the letters) is placed within its focal length. The magnification of a lens is inversely related to its focal length. A lens with a shorter focal length provides greater magnification. Therefore, a convex lens of focal length $5 \text{ cm}$ would be preferred over one with a focal length of $50 \text{ cm}$ for better magnification.

To obtain an erect image using a concave mirror, the object must be placed between the pole (P) and the principal focus (F) of the mirror.

Range of object distance: The focal length ($f$) is given as $15 \text{ cm}$. Therefore, the object distance ($u$) must be less than $15 \text{ cm}$. The range is $0 < u < 15 \text{ cm}$ from the pole of the mirror.

Nature of the image: The image formed is virtual and erect.

Size of the image: The image is larger than the object (magnified).

Ray Diagram: (See Figure 9.7 (f) in the textbook for a similar diagram.)

(a) Headlights of a car: Mirror type: Concave mirror. Reason: The light bulb is placed at the principal focus of the concave mirror. The mirror reflects the light rays from the bulb into a powerful, parallel beam, which illuminates the road over a long distance.

(b) Side/rear-view mirror of a vehicle: Mirror type: Convex mirror. Reason: A convex mirror is used because it provides a wider field of view than a plane mirror of the same size. It allows the driver to see a much larger area of the traffic behind. Also, it always forms an erect and diminished image of the objects.

(c) Solar furnace: Mirror type: Large concave mirror. Reason: A large concave mirror is used to concentrate a large amount of parallel sunlight onto a small area at its principal focus. This concentration of solar energy produces a very high temperature, which is used for heating and melting materials in the furnace.

Answer: Yes, the lens will still produce a complete image of the object.

Explanation: An image is formed when rays of light from a point on the object converge at a corresponding point after passing through the lens. Rays from every point on the object travel in all directions and strike the entire surface of the lens. Even if one-half of the lens is covered, light rays from the object can still pass through the other uncovered half and form a complete image at the same position.

However, the brightness or intensity of the image will be reduced. This is because the amount of light passing through the lens is halved, so fewer rays converge to form the image, making it dimmer.

Given: Height of the object, $h = +5 \text{ cm}$ Object distance, $u = -25 \text{ cm}$ (object is on the left) Focal length of converging (convex) lens, $f = +10 \text{ cm}$

To Find: Image distance, $v$ Height of the image, $h'$ Nature of the image

Formula: Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ Magnification formula: $m = \frac{h'}{h} = \frac{v}{u}$

Calculation: Using the lens formula: $\frac{1}{v} - \frac{1}{(-25)} = \frac{1}{10}$ $\frac{1}{v} + \frac{1}{25} = \frac{1}{10}$ $\frac{1}{v} = \frac{1}{10} - \frac{1}{25}$ $\frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50}$ $v = \frac{50}{3} \approx +16.67 \text{ cm}$

Now, calculating the size of the image using the magnification formula: $\frac{h'}{h} = \frac{v}{u}$ $\frac{h'}{5} = \frac{16.67}{-25}$ $h' = \frac{16.67 \times 5}{-25} = \frac{83.35}{-25} \approx -3.33 \text{ cm}$

Ray Diagram: The object is placed beyond $2F_1$ (since $f=10 \text{ cm}$, $2F_1 = 20 \text{ cm}$, and $u=25 \text{ cm}$). The diagram would show a real, inverted, and diminished image formed between $F_2$ and $2F_2$ on the other side of the lens.

Given: Focal length of concave lens, $f = -15 \text{ cm}$ Image distance, $v = -10 \text{ cm}$ (A concave lens always forms a virtual image on the same side as the object).

To Find: Object distance, $u$

Formula: Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Calculation: Substituting the given values into the lens formula: $\frac{1}{(-10)} - \frac{1}{u} = \frac{1}{(-15)}$ $-\frac{1}{10} - \frac{1}{u} = -\frac{1}{15}$ $-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}$ $-\frac{1}{u} = \frac{-2 + 3}{30} = \frac{1}{30}$ $\frac{1}{u} = -\frac{1}{30}$ $u = -30 \text{ cm}$

Final Answer: The object is placed at a distance of $30 \text{ cm}$ from the concave lens on the left side.

Ray Diagram: (See Figure 9.17 (b) in the textbook for a similar diagram.)

Given: Object distance, $u = -10 \text{ cm}$ Focal length of convex mirror, $f = +15 \text{ cm}$

To Find: Image distance, $v$ Nature of the image

Formula: Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Calculation: Substituting the values in the mirror formula: $\frac{1}{v} + \frac{1}{(-10)} = \frac{1}{15}$ $\frac{1}{v} - \frac{1}{10} = \frac{1}{15}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$ $\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30}$ $\frac{1}{v} = \frac{1}{6}$ $v = +6 \text{ cm}$

The magnification ($m$) produced by a plane mirror being $+1$ has two implications:

Therefore, $m = +1$ means that a plane mirror produces a virtual, erect image of the same size as the object.

Given: Height of the object, $h = +5.0 \text{ cm}$ Object distance, $u = -20 \text{ cm}$ Radius of curvature of convex mirror, $R = +30 \text{ cm}$

To Find: Image distance, $v$ Nature of the image Height of the image, $h'$

Formula: Focal length, $f = \frac{R}{2}$ Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ Magnification formula: $m = \frac{h'}{h} = -\frac{v}{u}$

Calculation: First, find the focal length: $f = \frac{+30}{2} = +15 \text{ cm}$

Now, use the mirror formula to find the image distance ($v$): $\frac{1}{v} + \frac{1}{(-20)} = \frac{1}{15}$ $\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$ $\frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60}$ $v = \frac{60}{7} \approx +8.57 \text{ cm}$

Now, calculate the size of the image ($h'$) using the magnification formula: $\frac{h'}{h} = -\frac{v}{u}$ $\frac{h'}{5.0} = -\frac{(60/7)}{(-20)}$ $h' = \frac{60}{7 \times 20} \times 5.0 = \frac{300}{140} = \frac{15}{7} \approx +2.14 \text{ cm}$

Given: Height of the object, $h = +7.0 \text{ cm}$ Object distance, $u = -27 \text{ cm}$ Focal length of concave mirror, $f = -18 \text{ cm}$

To Find: Image distance (screen position), $v$ Height of the image, $h'$ Nature of the image

Formula: Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ Magnification formula: $m = \frac{h'}{h} = -\frac{v}{u}$

Calculation: Using the mirror formula to find the image distance ($v$): $\frac{1}{v} + \frac{1}{(-27)} = \frac{1}{(-18)}$ $\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}$ $\frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54}$ $v = -54 \text{ cm}$

Now, calculate the size of the image ($h'$) using the magnification formula: $\frac{h'}{h} = -\frac{v}{u}$ $\frac{h'}{7.0} = -\frac{(-54)}{(-27)}$ $\frac{h'}{7.0} = -2$ $h' = -2 \times 7.0 = -14.0 \text{ cm}$

Given: Power of the lens, $P = -2.0 \text{ D}$ (Dioptre)

To Find: Focal length, $f$ Type of lens

Formula: Power of a lens, $P = \frac{1}{f \text{ (in metres)}}$

Calculation: $f = \frac{1}{P}$ $f = \frac{1}{-2.0} = -0.5 \text{ m}$ To convert to centimeters: $f = -0.5 \times 100 = -50 \text{ cm}$.

Given: Power of the lens, $P = +1.5 \text{ D}$ (Dioptre)

To Find: Focal length, $f$ Type of lens (diverging or converging)

Formula: Power of a lens, $P = \frac{1}{f \text{ (in metres)}}$

Calculation: $f = \frac{1}{P}$ $f = \frac{1}{+1.5} = +\frac{1}{1.5} = +\frac{10}{15} = +\frac{2}{3} \approx +0.67 \text{ m}$ To convert to centimeters: $f = +0.67 \times 100 = +67 \text{ cm}$.