Wave OpticsClass 12 Physics NCERT Solutions

Given:

First, let's calculate the frequency of the incident light. The frequency of light depends only on the source and does not change when the medium changes. Frequency, ν = c / λ_air ν = (3.0 × 10⁸ m/s) / (589 × 10⁻⁹ m) ν ≈ 5.09 × 10¹⁴ Hz

(a) For reflected light: The reflected light travels back into the same medium (air). Therefore, its properties remain the same as the incident light.

(b) For refracted light: The refracted light enters the water. Its speed and wavelength will change, but the frequency will remain the same.

A wavefront is defined as the locus of all points that are in the same phase of vibration.

(a) Light diverging from a point source. For a point source emitting light uniformly in all directions in an isotropic medium, the light waves travel outwards with the same speed in all directions. The locus of points at the same distance from the source will have the same phase. Therefore, the shape of the wavefront is a sphere with the point source at its center. These are called spherical wavefronts.

(b) Light emerging out of a convex lens when a point source is placed at its focus. When a point source of light is placed at the principal focus of a convex lens, the rays of light emerging from the lens become parallel to the principal axis. Since rays are always perpendicular to the wavefront, the wavefront for a parallel beam of light is a plane. Therefore, the emerging wavefront is a plane wavefront.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth. A distant star is effectively a point source at an infinite distance from the Earth. The light from such a source produces spherical wavefronts. However, because the distance is extremely large, the radius of the spherical wavefront is also extremely large. The small portion of this large spherical wavefront that is intercepted by the Earth has a very small curvature and can be approximated as a plane. Therefore, the shape of the wavefront is a plane wavefront.

Given:

The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v).

n = c / v

Rearranging for the speed of light in glass (v_g):

v_g = c / n_g v_g = (3.0 × 10⁸ m/s) / 1.5 v_g = 2.0 × 10⁸ m/s

So, the speed of light in glass is 2.0 × 10⁸ m/s.

No, the speed of light in glass is not independent of the colour of light. The refractive index of a material like glass depends on the wavelength (and hence colour) of the light passing through it. This phenomenon is known as dispersion.

According to Cauchy's relation, the refractive index (n) is generally greater for shorter wavelengths. In the visible spectrum, violet light has a shorter wavelength than red light.

λ_violet < λ_red

Therefore, the refractive index of glass for violet light is greater than that for red light.

n_violet > n_red

Since the speed of light in the medium is v = c/n, the speed is inversely proportional to the refractive index. A higher refractive index means a lower speed.

As n_violet > n_red, it follows that v_violet < v_red.

Therefore, violet light travels slower than red light in a glass prism.

Given:

In a Young's double-slit experiment, the position (x_n) of the n-th bright fringe from the central maximum is given by the formula:

x_n = nλD / d

where λ is the wavelength of the light.

For the central bright fringe, n = 0, so its position is x_0 = 0. For the fourth bright fringe, n = 4, so its position is:

x_4 = 4λD / d

The distance between the central bright fringe and the fourth bright fringe is:

Δx = x_4 - x_0 = (4λD / d) - 0 = 4λD / d

We are given that this distance is 1.2 cm.

1.2 × 10⁻² m = 4λD / d

We can rearrange this equation to solve for the wavelength λ:

λ = (Δx × d) / (4 × D)

Substituting the given values:

λ = (1.2 × 10⁻² m × 0.28 × 10⁻³ m) / (4 × 1.4 m) λ = (0.336 × 10⁻⁵ m²) / 5.6 m λ = 0.06 × 10⁻⁵ m λ = 6.0 × 10⁻⁷ m

Converting this to nanometers (1 nm = 10⁻⁹ m):

λ = 600 × 10⁻⁹ m = 600 nm

Therefore, the wavelength of the light used in the experiment is 600 nm.

The intensity (I) at a point on the screen in a Young's double-slit experiment is given by:

I = 4I₀ cos²(φ/2)

where I₀ is the intensity from a single slit and φ is the phase difference between the waves from the two slits.

The phase difference (φ) is related to the path difference (Δx) by the formula:

φ = (2π/λ) × Δx

Case 1: Path difference is λ Given path difference, Δx = λ. Let's find the phase difference: φ₁ = (2π/λ) × λ = 2π

The intensity at this point is I₁: I₁ = 4I₀ cos²(φ₁/2) = 4I₀ cos²(2π/2) = 4I₀ cos²(π) Since cos(π) = -1, cos²(π) = 1. I₁ = 4I₀(1) = 4I₀ We are given that this intensity is K units. So, K = 4I₀.

Case 2: Path difference is λ/3 Given path difference, Δx' = λ/3. Let's find the new phase difference: φ₂ = (2π/λ) × (λ/3) = 2π/3

The intensity at this new point is I₂: I₂ = 4I₀ cos²(φ₂/2) = 4I₀ cos²((2π/3)/2) = 4I₀ cos²(π/3) We know that cos(π/3) = cos(60°) = 1/2. I₂ = 4I₀ (1/2)² = 4I₀ (1/4) = I₀

Now, we need to express I₂ in terms of K. From Case 1, we know that K = 4I₀, which means I₀ = K/4.

Substituting this into the expression for I₂:

I₂ = I₀ = K/4

Therefore, the intensity of light at a point where the path difference is λ/3 is K/4 units.

Given:

For this problem, we will assume the experimental setup is the same as in question 10.4:

The position of the n-th bright fringe is given by x_n = nλD/d.

(a) Distance of the third bright fringe for λ₁ = 650 nm Here, n = 3 and λ = λ₁.

x₃ = (3 × λ₁ × D) / d x₃ = (3 × 650 × 10⁻⁹ m × 1.4 m) / (0.28 × 10⁻³ m) x₃ = (2730 × 10⁻⁹) / (0.28 × 10⁻³) m x₃ = 9750 × 10⁻⁶ m = 9.75 × 10⁻³ m x₃ = 0.975 cm

The distance of the third bright fringe for wavelength 650 nm is 0.975 cm.

(b) Least distance for coincidence of bright fringes For the bright fringes of the two wavelengths to coincide, their distances from the central maximum must be equal. Let the n₁-th bright fringe of λ₁ coincide with the n₂-th bright fringe of λ₂.

x = (n₁λ₁D) / d = (n₂λ₂D) / d

This simplifies to:

n₁λ₁ = n₂λ₂ n₁ / n₂ = λ₂ / λ₁ n₁ / n₂ = (520 nm) / (650 nm) = 52/65 = 4/5

We are looking for the least distance from the central maximum, so we need the smallest integer values for n₁ and n₂ that satisfy this ratio. The smallest integers are n₁ = 4 and n₂ = 5.

This means the 4th bright fringe of the 650 nm light coincides with the 5th bright fringe of the 520 nm light.

We can find the distance (x_coincide) using either wavelength:

Using λ₁: x_coincide = (n₁λ₁D) / d x_coincide = (4 × 650 × 10⁻⁹ m × 1.4 m) / (0.28 × 10⁻³ m) x_coincide = (3640 × 10⁻⁹) / (0.28 × 10⁻³) m x_coincide = 13000 × 10⁻⁶ m = 1.3 × 10⁻² m x_coincide = 1.3 cm

Using λ₂ (to verify): x_coincide = (n₂λ₂D) / d x_coincide = (5 × 520 × 10⁻⁹ m × 1.4 m) / (0.28 × 10⁻³ m) x_coincide = (3640 × 10⁻⁹) / (0.28 × 10⁻³) m x_coincide = 1.3 cm

The least distance from the central maximum where the bright fringes coincide is 1.3 cm.