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Physics · 2026 · 5 marks
Ray Optics and Optical InstrumentsRefraction at Spherical Surfaces and by Lenses5 marksApplylong answer
CBSE 2026 · Region 1 · Set 1 · Q32
(a) Using the relation for refraction at a curved spherical surface, derive the expression for lens maker's formula. (b) Three lenses L_1, L_2 and…
(a)
Using the relation for refraction at a curved spherical surface, derive the expression for lens maker's formula.
(b)
Three lenses $\displaystyle \mathrm{L}_{1}, \mathrm{~L}_{2}$ and $\displaystyle \mathrm{L}_{3}$, each of focal length $\displaystyle 40$ cm, are placed coaxially. The distance between $\displaystyle \mathrm{L}_{1}$ and $\displaystyle \mathrm{L}_{2}$ and between $\displaystyle \mathrm{L}_{2}$ and $\displaystyle \mathrm{L}_{3}$ are $\displaystyle 120$ cm and $\displaystyle 20$ cm respectively. An object is kept at a distance of $\displaystyle 80$ cm to the left of lens $\displaystyle \mathrm{L}_{1}$. Find the distance of the final image formed from the object.
(a)
Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the centre of curvature. Using this diagram, derive the mirror formula.
(b)
A concave mirror produces a two times magnified virtual image of an object kept $\displaystyle 10$ cm in front of it. Calculate the focal length of the mirror.
Marking-scheme solution
(a)
The first refracting surface forms the image of the object O at $\displaystyle \mathrm{I}_{1}$
For refraction from first interface ABC
$\displaystyle \dfrac{\mathrm{n}_{1}}{\mathrm{OB}}+\dfrac{\mathrm{n}_{2}}{\mathrm{BI}_{1}}=\dfrac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{BC}_{1}}$ ---------- ($\displaystyle 1$)
Similarly for refraction from second interface ADC. The image $\displaystyle \mathrm{I}_{1}$ acts as a virtual object for the second surface ADC
$\displaystyle -\dfrac{\mathrm{n}_{2}}{\mathrm{DI}_{1}}+\dfrac{\mathrm{n}_{1}}{\mathrm{DI}}=\dfrac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{DC}_{2}}$ ---------- ($\displaystyle 2$)
For thin lens $\displaystyle \mathrm{BI}_{1}=\mathrm{DI}_{1}$ By adding equation ($\displaystyle 1$) and ($\displaystyle 2$)
$\displaystyle \dfrac{\mathrm{n}_{1}}{OB}+\dfrac{\mathrm{n}_{1}}{D \mathrm{I}}=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)\left[\dfrac{1}{BC_{1}}+\dfrac{1}{DC_{2}}\right]$
$\displaystyle -\dfrac{\mathrm{n}_{1}}{u}+\dfrac{\mathrm{n}_{1}}{v}=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]$
$\displaystyle \dfrac{1}{v}-\dfrac{1}{u}=\left(\dfrac{\mathrm{n}_{2}}{\mathrm{n}_{1}}-1\right)\left[\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right]$
If object is kept at $\displaystyle \infty$, image will form at focus hence
$\displaystyle \dfrac{1}{f}=\left(\dfrac{\mathrm{n}_{2}}{\mathrm{n}_{1}}-1\right)\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)$
(b)
For lens $\displaystyle \mathrm{L}_{1}$
$\displaystyle \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
$\displaystyle \dfrac{1}{40}=\dfrac{1}{v_{1}}+\dfrac{1}{80}$
$\displaystyle \dfrac{1}{v_{1}}=\dfrac{1}{80}$
$\displaystyle v_{1}=80 \mathrm{~cm}$
For lens $\displaystyle \mathrm{L}_{2}$
$\displaystyle u_{2}=120-80$
$\displaystyle u_{2}=40 \mathrm{~cm}$
$\displaystyle \dfrac{1}{40}=\dfrac{1}{v_{2}}+\dfrac{1}{40}$
$\displaystyle \dfrac{1}{v_{2}}=0$
$\displaystyle v_{2}=\infty$
For lens $\displaystyle \mathrm{L}_{3}$
$\displaystyle u_{3}=\infty$
$\displaystyle \dfrac{1}{40}=\dfrac{1}{v_{3}}+\dfrac{1}{\infty}$
$\displaystyle v_{3}=40 \mathrm{~cm}$
Distance between final image and object
$\displaystyle =80+120+20+40$
$\displaystyle =260 \mathrm{~cm}$
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