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Physics · 2025 · 5 marks
Ray Optics and Optical InstrumentsRefraction at Spherical Surfaces and by Lenses5 marksApplylong answer
CBSE 2025 · Region 4 · Set 2 · Q32
(a) (i) An object is placed 30 cm from a thin convex lens of focal length 10 cm. The lens forms a sharp image on a screen. If a thin concave lens is…
(i)
An object is placed $\displaystyle 30$ cm from a thin convex lens of focal length $\displaystyle 10$ cm. The lens forms a sharp image on a screen. If a thin concave lens is placed in contact with the convex lens, the sharp image on the screen is formed when the screen is moved by $\displaystyle 45$ cm from its initial position. Calculate the focal length of the concave lens.
(ii)
Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is $\displaystyle \sqrt{3}$. Calculate the angle of incidence for this case of minimum deviation also.
(i)
A physics teacher wants to demonstrate interference with the help of double slit experiment using a laser beam of $\displaystyle 633$ nm wavelength. Since the hall is large enough, interference pattern is formed on the wall $\displaystyle 5.0$ m from the slits. For clear and comfortable view by all the students they want the fringe width $\displaystyle 5$ mm.
(I)
Find the slit separation for obtaining the desired interference pattern.
(II)
How far will the first minimum be from the central maximum?
(ii)
A parallel beam of light of wavelength $\displaystyle 650$ nm passes through a slit of width $\displaystyle 0.6$ mm. The diffraction pattern is obtained on a screen kept $\displaystyle 60$ cm away from the slit. Find the distance between first order minima on both sides of the central maximum.
Marking-scheme solution
(i)
For real image formed by convex lens:
$\displaystyle \frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u_{1}}$
$\displaystyle \frac{1}{10}=\frac{1}{v_{1}}-\frac{1}{(-30)}$
$\displaystyle v_{1}=15 \mathrm{~cm}$
For the combination of lenses, let the focal length of the combination be $\displaystyle f_{3}$:
$\displaystyle \frac{1}{f_{3}}=\frac{1}{v_{3}}-\frac{1}{u_{3}}$
$\displaystyle \frac{1}{f_{3}}=\frac{1}{(15+45)}+\frac{1}{30}$
$\displaystyle f_{3}=20 \mathrm{~cm}$
Let the focal length of the concave lens be $\displaystyle f_{2}$:
$\displaystyle \frac{1}{f_{3}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$\displaystyle \frac{1}{f_{2}}=\frac{1}{20}-\frac{1}{10}$
$\displaystyle f_{2}=-20 \mathrm{~cm}$
(ii)
Angle of minimum deviation:
$\displaystyle \mu=\frac{\sin \dfrac{A+\delta_{m}}{2}}{\sin \dfrac{A}{2}}$
$\displaystyle \sqrt{3}=\frac{\sin \dfrac{60^{\circ}+\delta_{m}}{2}}{\sin 30^{\circ}}$
$\displaystyle \frac{\sqrt{3}}{2}=\sin \frac{A+\delta_{m}}{2}$
$\displaystyle 60^{\circ}=\frac{A+\delta_{m}}{2}$
$\displaystyle \delta_{m}=60^{\circ}$
Angle of incidence:
$\displaystyle i+e=A+\delta$
$\displaystyle 2 i=A+\delta_{m}$
$\displaystyle i=\frac{A+\delta_{m}}{2}$
$\displaystyle i=60^{\circ}$
(i)
(I)
Slit separation:
$\displaystyle \beta=\frac{D \lambda}{d}$
$\displaystyle d=\frac{D \lambda}{\beta}$
$\displaystyle =\frac{633 \times 10^{-9} \times 5}{5 \times 10^{-3}}$
$\displaystyle =633 \times 10^{-6} \mathrm{~m}=633 \mu \mathrm{m}$
(II)
Distance of first minimum from central maximum:
$\displaystyle x_{n}=\frac{(2 n-1) \lambda D}{2 d}$
$\displaystyle n=1$
$\displaystyle x=\frac{633 \times 10^{-9} \times 5}{2 \times 5 \times 10^{-3}}$
$\displaystyle x=316.5 \times 10^{-6} \mathrm{~m}=316.5 \mu \mathrm{m}$
(ii)
Distance between first order minima on both sides:
$\displaystyle W=\frac{2 D \lambda}{d}$
$\displaystyle =\frac{2 \times 650 \times 10^{-9}}{0.6 \times 10^{-3}} \times 60 \times 10^{-2}$
$\displaystyle =1.3 \times 10^{-3} \mathrm{~m}$
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