Question 48: Ray Optics

Question

List I contains four combinations of two lenses (1 and 2) whose focal lengths (in $\mathrm{cm}$ ) are indicated in the figures. In all cases, the object is placed $20 \mathrm{~cm}$ from the first lens on the left, and the distance between the two lenses is $5 \mathrm{~cm}$. List II contains the positions of the final images.

List-I
List-II

(I) 
(P) Final image is formed at $7.5 \mathrm{~cm}$ on the right side of lens 2 .

(II) 
(Q) Final image is formed at $60.0 \mathrm{~cm}$ on the right side of lens 2 .

(III) 
(R) Final image is formed at $30.0 \mathrm{~cm}$ on the left side of lens $2 .$

(IV) 
(S) Final image is formed at $6.0 \mathrm{~cm}$ on the right side of lens 2 .

(T) Final image is formed at $30.0 \mathrm{~cm}$ on the right side of lens 2 .

Which one of the following options is correct?

Accepted Answer

(I) $$u =  - 20$$ cm
$$f =  + 10$$ cm
$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$
$${1 \over v} = {1 \over {10}} - {1 \over {20}}$$
$${1 \over v} = {1 \over {20}}$$
$$v = 20$$ cm
$$u =  + 15$$ cm
$$f =  + 15$$ cm
$${1 \over v} - {1 \over u} = {1 \over f}$$
$$ \Rightarrow {1 \over v} - {1 \over {15}} = {1 \over {15}}$$
$${1 \over v} = {2 \over {15}}$$
$$v = 7.5$$ cm (from lens 2)
I $$	o$$ P
(II) $$u =  - 20$$ cm, $$f =  + 10$$ cm
$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$
$${1 \over v} = {1 \over {10}} - {1 \over {20}} \Rightarrow v = 20$$ cm
$$u =  + 15$$ cm
$$f =  - 10$$ cm
$${1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over v} - {1 \over {15}} = {{ - 1} \over {10}}$$
$${1 \over v} = {{ - 1} \over {10}} + {1 \over {15}} = {{ - 3 + 2} \over {30}}$$
$${1 \over v} =  - {1 \over {30}}$$
$$v =  - 30$$ cm
II $$	o$$ R
(III) $$u =  - 20$$ cm
$$f =  + 10$$ cm
$${1 \over v} + {1 \over {20}} = {1 \over {10}}$$
$$\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$$
$$v = 20$$ cm
$$ \Rightarrow u = 15$$ cm
$$f =  - 20$$ cm
$${1 \over v} - {1 \over {15}} = {{ - 1} \over {20}}$$
$${1 \over v} =  - {1 \over {20}} + {1 \over {15}} = {{ - 3 + 4} \over {60}} = {1 \over {60}}$$
$$v = 60$$ cm
III $$	o$$ Q
(IV) $$u =  - 20$$ cm
$$f =  - 20$$ cm
$$ \Rightarrow {1 \over v} - {1 \over u} = {1 \over f}$$
$$ \Rightarrow {1 \over v} + {1 \over {20}} = {{ - 1} \over {20}}$$
$$v =  - 10$$ cm
$$u =  - 15$$ cm
$$f = 10$$ cm
$${1 \over v} - {1 \over u} = {1 \over f}$$
$${1 \over v} + {1 \over {15}} = {1 \over {10}}$$
$$ \Rightarrow {1 \over v} = {1 \over {10}} - {1 \over {15}}$$
$$ = {{3 - 2} \over {30}} = {1 \over {30}}$$
$$v = 30$$ cm
IV $$	o$$ T

Answer

(I) $\rightarrow$ P; (II) $\rightarrow$ R; (III) $\rightarrow$ Q; (IV) $\rightarrow$ T

Answer

(I) $\rightarrow$ Q; (II) $\rightarrow$ P; (III) $\rightarrow$ T; (IV) $\rightarrow$ S

Answer

(I) $\rightarrow \mathrm{P}$; (II) $\rightarrow \mathrm{T}$; (III) $\rightarrow \mathrm{R}$; (IV) $\rightarrow \mathrm{Q}$

Answer

(I) $\rightarrow$ T; (II) $\rightarrow$ S; (III) $\rightarrow$ Q; (IV) $\rightarrow$ R

Ray Optics

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