JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Geometrical Optics

Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is

r

and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.

List I	List II
P. (see structure)	1. $2 r$
Q. (see structure)	2. $r /2$
R. (see structure)	3. $- r$
S. (see structure)	4. $r$

(A)P-1, Q-2, R-3, S-4
P
(B)P-2, Q-4, R-3, S-1
Q
✓ Correct
(C)P-4, Q-1, R-2, S-3
R
(D)P-2, Q-1, R-3, S-4
S

Answer: (B)

Solution

Using the lensmaker's equation with

μ = 1.5

Biconvex lens: $f = \frac{r}{2 ( μ - 1 )} = r$
Plano-convex lens: $f = \frac{r}{μ - 1} = 2 r$
Plano-concave lens: $f = \frac{- r}{μ - 1} = - 2 r$

P (two biconvex in contact):

\frac{1}{f _{e q}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}

, so

f_{e q} = r /2

. P→2.

Q (two plano-convex in contact):

\frac{1}{f _{e q}} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{1}{r}

, so

f_{e q} = r

. Q→4.

R (two plano-concave in contact):

\frac{1}{f _{e q}} = \frac{- 1}{2 r} + \frac{- 1}{2 r} = \frac{- 1}{r}

, so

f_{e q} = - r

. R→3.

S (biconvex + plano-concave):

\frac{1}{f _{e q}} = \frac{1}{r} - \frac{1}{2 r} = \frac{1}{2 r}

, so

f_{e q} = 2 r

. S→1.

Answer: P-2, Q-4, R-3, S-1.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

Practise more: all JEE Advanced PYQs on Geometrical Optics →

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Geometrical Optics

Solution

More JEE Advanced questions on Geometrical Optics