Solution-Part-2-Class-12-Physics-Chapter-7-Wave Optics-MSBSHSE

Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively.

Let λ_v be the wavelength of light in vacuum.

λ₁ = λ_v/n = \(\frac{400×10^{-9}}{1.55}\)  = 258.06 x 10⁻⁹ m

λ₂ = λ_v/n = \(\frac{700×10^{-9}}{1.55}\) = 451.61 x 10⁻⁹ m

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Let d_fg and d_m be the distances by the ray of light in the flint glass and the medium respectively. Also, let n_fg and n_m be the refractive indices of the flint glass and the medium respectively.

Given : d_fg = 3 cm, n_fg = 1.6, n_m = 1.25,

Optical path = n_m x d_m = n_fg x d_fg

∴ d_m = \(\frac{n_{fg}×d_{fg}}{n_m}\) = 3.84 cm

Thus, x cm = 3.84 cm

∴ x = 3.84

This is the value of x.

Given : θ₁ = 0.20°, n_w = 1.33

In the first approximation,

D sin θ₁ = y₁ and D sin θ₂ = y₂

sin θ₂/sin θ₁ = y₂/y₁   …..(1)

Now, y ∝ λD/d

For given d and D,

y ∝ λ

∴ y₂/y₁ = λ₂/λ₁   …..(2)

Now, n_w = λ₁/λ₂     …..(3)

From Eqs. (1), (2) and (3), we get,

sin θ₂/sin θ₁ = λ₂/λ₁ = 1/n_w

∴ sin θ₂ = sin θ₁/n_w = \(\frac{sin0.2}{1.33}=\frac{0.0035}{1.33}\)  = 0.0026

θ₂ = sin^-10.0026 = 9' = 0.15°

This is the required angular fringe separation

Data : d = 100 λ, D = 50.0 cm

a) The condition for maximum intensity in Young's experiment is,

d sin θ = nλ , n = 0, 1, 2...,

The angle between the central maximum and itsadjacent maximum can be determined by setting n equal to 1,

∴ d sin θ = λ

∴ θ = sin⁻¹(λ/d )= sin⁻¹(λ/100λ) = sin⁻¹(1/100)  = 0.9’ = 0.01571 rad

(b) The distance between these maxima on the screen is

D sin θ = D(λ/d) = 50(λ/100λ) = 0.50 cm

According to Malus’ law, when the unpolarized light with intensity I₀ is incident on the first polarizer, the polarizer polarizes this incident light. The intensity of light becomes I₁ = I₀/2.

Now, I₂ = I₁ cos²θ

I₂ = \((\frac{I_0}{2})\) cos²θ

Also, the angle θ between the axes of the two polarizers is θ₂ — θ₁.

∴    I₂ = \((\frac{I_0}{2})\)cos²(θ₂ — θ₁)

=  \((\frac{I_0}{2})\)cos²(90° — 50°)

I₂  = \((\frac{I_0}{2})\)cos²40°

The intensity of light after it has passed through the second polaroid  =  \((\frac{I_0}{2})\)cos²40° = \((\frac{I_0}{2})\)(0.7660)²

= 0.2934 I₀

Given : D = 1.2 m

The distance between the central bright band and the 20th bright band is 0.4 cm.

∴ y₂₀  = 0.4 cm = 0.4 x 10⁻² m

W = y₂₀/20 = = (0.4 x 10⁻²)/20 = = 2 x 10⁻⁴ m, d₁ = 0.9 cm = 0.9 x 10⁻² m,

v₁ = 90 cm = 0.9  m

∴ u₁ = D − v₁ = 1.2 − 0.9 = 0.3 m

Now d₁/d = v₁/u₁

∴ d = \(\frac{d_1u_1}{v_1}=\frac{0.9×10^{-2}×0.3}{0.9}\) = 3 x 10⁻³ m,

∴ The wavelength of light,

λ = \(\frac{Wd}{D}=\frac{2×10^{-4}×3×10^{-3}}{1.2}\)m

= 5 x 10⁻⁷ m,

= 5 x 10⁻⁷ x 10¹⁰ Å

= 5000 Å

Given: D = 2 m, y_1d = 5 mm = 5 x 10⁻³ m, a = 0.2 mm = 0.2 x 10⁻³ m  = 2 x 10⁻⁴ m

y_md = \(m\frac{λD}{a}\)

∴ λ = \(\frac{y_{1d}a}{D}=\frac{5×10^{-3}×0.2×10^{-3}}{2}\)

∴ λ = 5 x 10⁻⁷ m = 5 x 10⁻⁷ x 10⁻¹⁰ Å = 5000 Å

This is the wavelength of light.

Given : I₁:I₂ = 2:1

If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

\(\frac{I_{max}}{i_{min}}=(\frac{E_{10}+E_{20}}{E_{10}-E_{20}})^2 =(\frac{r+1}{r-1})^2\)

where r = E₁₀/E₂₀.

∴ \(\frac{I_1}{I_2}=(\frac{E_{10}}{E_{20}})^2=r^2\)

∴ r =\(\sqrt{\frac{I_1}{I_2}}\)

∴ \(\frac{I_{max}}{i_{min}}=(\frac{\sqrt{2}+1}{\sqrt{2}-1})^2=(\frac{2.414}{0.414})^2\) = (5.83)² = 33.99 = 34

The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.

Given : λ = 546 nm = 546 x 10⁻⁹ m, a = 0.4 mm = 4 x 10⁻⁴ m,

D = 40 cm = 40 x 10⁻² m,

y_md =  \(m\frac{λD}{a}\)

∴ y_1d = \(1\frac{λD}{a}\)

And 2y_1d = \(2\frac{λD}{a}\)=(\frac{2×546×10^{-9}}{4×10^{-4}}\)

= 2 x 546 x 10⁻⁶ = 1092 x 10⁻⁶

= 1.092 x 10⁻³ = 1.1 mm

This is the distance between the two first order minima.

Given : θ = 45°, m = 1

 a sin θ = mλ for (m = 1, 2, 3... minima)

Here, m = 1 (First minimum)

 a sin 45° = (1) λ

∴ a/λ = 1/sin 45° = 1.414

This is the required ratio.

Given : 2W = 6mm W = 3mm = 3 x 10⁻³ m, y=2.5 m,

(a) λ₁ = 500 nm = 5 x 10⁻⁷ m

(b) λ₂ = 50 μm = 5 x 10⁻⁵ m

(c) λ₃ = 0.500 nm = 5 x 10⁻¹⁰ m

Let a be the slit width.

(a) W = \(\frac{yλ_1}{a}\)

∴ a = \(\frac{yλ_1}{W}\) = \(\frac{(2.5)(5×10^{-7})}{3×10^{-3}}\)

= 4.167 x 10⁻⁴ m

= 0.4167 mm

(b) W =\(\frac{yλ_2}{a}\)

∴ a = \(\frac{yλ_2}{W}\)=\(\frac{(2.5)(5×10^{-10})}{3×10^{-3}}\)

= 4.167 x 10⁻² m

= 41.67 mm

(C) W = \(\frac{yλ_3}{a}\)

∴ a = \(\frac{yλ_3}{W}\)=\(\frac{(2.5)(5×10^{-5})}{3×10^{-3}}\)

= 4.167 x 10⁻⁷ m

= 4.167 x 10⁻⁴ mm

Given : λ = 5000 Å = 5 x 10⁻⁷ m, D = 200 x 2.54 cm = 5.08 m

θ = 1.22λ/D  =  \(\frac{(1.22)(5×10^{-7})}{5.08}\)

= 1.2×10^-7 rad

This is the required quantity

Given : λ₁ = 6000 Å = 6 x 10⁻⁷ m, λ₂ = 4800 Å = 4.8 x 10⁻⁷ m,

W₁ = 0.32 mm =3.2 x 10⁻⁴ m

Distance between consecutive bright fringes,

W = \(\frac{λD}{d}\)

For λ₁  W₁ = \(\frac{λ_1D}{d}\)    ……. (1) and

For λ₂  W₂ = \(\frac{λ_2D}{d}\)   ……. (2)

\(\frac{W_2}{W-1} = \frac{λ_2D/d}{λ_1D/d}\)

∴ W₂ = \(\frac{λ_2}{λ_1}W_1\) = \(\frac{4.8×10^{-7}}{6×10^{-7}}(3.2×10^{-4})\)

        = 0.8 x 3.2 x 10⁻⁴

        = 2.56 x 10⁻⁴

∴ ΔW = W₁ − W₂ = 3.2 x 10⁻⁴ − 2.56 x 10⁻⁴ =0.64 x 10⁻⁴ m

 = 0.064 mm

This is the required change in distance.