JEE Advance - Physics (2014 - Paper 2 Offline - No. 19)
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.
Giải thích
The lens maker's formula gives focal length of the bi-convex lens as
$${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$$
$$ = (1.5 - 1)\left[ {{1 \over r} - {1 \over { - r}}} \right] = {1 \over r}$$ ..... (1)
The focal length of a plano-convex (also convex-plane) lens is given by
$${1 \over {{f_2}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over { - r}}} \right] = {1 \over {2r}}$$ ...... (2)
and that of a plano-concave (also concave-plano) lens is given by
$${1 \over {{f_3}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over r}} \right] = - {1 \over {2r}}$$ ..... (3)
Case P:
Therefore, $${1 \over {{f_{eff}}}} = {1 \over r} + {1 \over r}$$
$${f_{eff}} = {r \over 2}$$
(P) $$\to$$ (2)
Case Q :
$${1 \over {{f_{eff}}}} = {1 \over {2r}} + {1 \over {2r}}$$
$${f_{eff}} = r$$
(Q) $$\to$$ (4)
Case R :
$${1 \over {{f_{eff}}}} = - {1 \over {2r}} - {1 \over {2r}}$$
$${f_{eff}} = - r$$
(R) $$\to$$ (3)
Case S : $${1 \over {{f_{eff}}}} = {1 \over r} - {1 \over {2r}} = {1 \over {2r}}$$
$${f_{eff}} = 2r$$
(S) $$\to$$ (1)