Wave Optics
Maharashtra BoardClass12thPhysicsChapter7
SolutionPart1
Solution : Part1
Solution : Part2

Question 13.
White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55?
Let λ_{1} and λ_{2} 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.
λ_{1} = λ_{v}/n = \(\frac{400×10^{9}}{1.55}\) = 258.06 x 10^{−}^{9} m
λ_{2} = 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.
Question 14.
The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.
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.
Question 15.
A doubleslit arrangement produces interference fringes for sodium light (λ =589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?
Given : θ_{1} = 0.20°, n_{w} = 1.33
In the first approximation,
D sin θ_{1 }= y_{1} and D sin θ_{2 }= y_{2}
sin θ_{2}/sin θ_{1} = y_{2}/y_{1} …..(1)
Now, y ∝ λD/d
For given d and D,
y ∝ λ
∴ y_{2}/y_{1} = λ_{2}/λ_{1} …..(2)
Now, n_{w} = λ_{1}/λ_{2} …..(3)
From Eqs. (1), (2) and (3), we get,
sin θ_{2}/sin θ_{1} = λ_{2}/λ_{1} = 1/n_{w}
∴ sin θ_{2} = sin θ_{1}/n_{w} = \(\frac{sin0.2}{1.33}=\frac{0.0035}{1.33}\) = 0.0026
θ_{2} = sin^{1}0.0026 = 9' = 0.15°
This is the required angular fringe separation
Question 16.
In a doubleslit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum?
(b) What is the distance between these maxima on a screen 50.0 cm from the slits?
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
Question 17.
Unpolarized light with intensity I_{0} is incident on two polaroids. The axis of the first polaroid makes an angle of 50^{o} with the vertical, and the axis of the second polaroid is horizontal. What is the intensity of the light after it has passed through the second polaroid?
According to Malus’ law, when the unpolarized light with intensity I_{0} is incident on the ﬁrst polarizer, the polarizer polarizes this incident light. The intensity of light becomes I_{1} = I_{0}/2.
Now, I_{2} = I_{1} cos^{2}θ
I_{2} = \((\frac{I_0}{2})\) cos^{2}θ
Also, the angle θ between the axes of the two polarizers is θ_{2} — θ_{1}.
∴ I_{2} = \((\frac{I_0}{2})\)cos^{2}(θ_{2} — θ_{1})
= \((\frac{I_0}{2})\)cos^{2}(90° — 50°)
I_{2} = \((\frac{I_0}{2})\)cos^{2}40°
The intensity of light after it has passed through the second polaroid = \((\frac{I_0}{2})\)cos^{2}40° = \((\frac{I_0}{2})\)(0.7660)^{2}
= 0.2934 I_{0}
Question 18.
In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.
Given : D = 1.2 m
The distance between the central bright band and the 20th bright band is 0.4 cm.
∴ y_{20} = 0.4 cm = 0.4 x 10^{−}^{2} m
W = y_{20}/20 = = (0.4 x 10^{−}^{2})/20 = = 2 x 10^{−}^{4} m, d_{1} = 0.9 cm = 0.9 x 10^{−}^{2} m,
v_{1} = 90 cm = 0.9 m
∴ u_{1} = D − v_{1} = 1.2 − 0.9 = 0.3 m
Now d_{1}/d = v_{1}/u_{1}
∴ 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 10^{10} Å
= 5000 Å
Question 19.
In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light.
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.
Question 20.
The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern?
Given : I_{1}:I_{2} = 2:1
If E_{10} and E_{20} are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is
\(\frac{I_{max}}{i_{min}}=(\frac{E_{10}+E_{20}}{E_{10}E_{20}})^2 =(\frac{r+1}{r1})^2\)
where r = E_{10}/E_{20}.
∴ \(\frac{I_1}{I_2}=(\frac{E_{10}}{E_{20}})^2=r^2\)
∴ 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.
Question 21.
A parallel beam of green light of wavelength 546 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?
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.
Question 22.
What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°?
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.
Question 23.
Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is (a) 500 nm (visible light); (b) 50 μm (infrared radiation); (c) 0.500 nm (Xrays)?
Given : 2W = 6mm W = 3mm = 3 x 10^{−}^{3 }m, y=2.5 m,
(a) λ_{1} = 500 nm = 5 x 10^{−}^{7 }m
(b) λ_{2} = 50 μm = 5 x 10^{−}^{5 }m
(c) λ_{3 }= 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
Question 24.
A star is emitting light at the wavelength of 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch.
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
Question 25.
The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?
Given : λ_{1} = 6000 Å = 6 x 10^{−}^{7} m, λ_{2 }= 4800 Å = 4.8 x 10^{−}^{7} m,
W_{1} = 0.32 mm =3.2 x 10^{−}^{4} m
Distance between consecutive bright fringes,
W = \(\frac{λD}{d}\)
For λ_{1} W_{1} = \(\frac{λ_1D}{d}\) ……. (1) and
For λ_{2} W_{2} = \(\frac{λ_2D}{d}\) ……. (2)
\(\frac{W_2}{W1} = \frac{λ_2D/d}{λ_1D/d}\)
∴ W_{2} = \(\frac{λ_2}{λ_1}W_1\) = \(\frac{4.8×10^{7}}{6×10^{7}}(3.2×10^{4})\)
= 0.8 x 3.2 x 10^{−}^{4}
= 2.56 x 10^{−}^{4}
∴ ΔW = W_{1} − W_{2 }= 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.
Main Page : – Maharashtra Board Class 12thPhysics – All chapters notes, solutions, videos, test, pdf.
Previous Chapter : Chapter6Superposition of Waves –Online Solution
Next Chapter : Chapter8 Electrostatics – Online Solution
We reply to valid query.