A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical directiion by 0.6 cm as shown . The distance between the lens and mirror is 30 cm . An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens . if A'B' is the image after refraction from the lens and the reflectiion from the mirror , find the distance of A'B' from the pole of the mirror and obtain its magnification . Also locate positions of A' and B' with respect to the optic axis RS.

Answer

A, B, C

Explanation

(a) For the lens
`(1)/(v)-(1)/(u)=(1)/(f_l)implies(1)/(v)-(1)/(20)=(1)/(15)`
`implies(1)/(v)=(1)/(15)-(1)/(20)=(1)/(60)impliesv=60`cm, `m=(v)/(u)=(60)/(-20)=-3`
The image is formed to the left of the lens, real inverted and three times the actural size (3.6cm in height below PQ).
For the mirror,
`(1)/(v^`)+(1)/(u^`)=(1)/(f_m)implies(1)/(v^`)=(1)/(-30)-(1)/(30)=-(2)/(30)`
`impliesv^`=-15`cm
`m=-(v^`)/(u^`)=-(-15)/(30)=(1)/(2)`
size of image`=(1)/(2)xx3.6=1.8`cm
This image will be inverted w.r.t. the original image and its position will be 0.3cm above RS and 1.5cm below RS. The position of the image is 15cm to the right of the mirror.
(b) The path difference between the two rays reflected from the upper surface AB (shown by ray 1, single arroe upwards) and lower surface CD (shown by ray 2 double arrow pointing upwards) is
`trianglex=_mnxx2t+(lamda)/(2)`
Here `(lamda)/(2)` is the path difference as the ray 1 suffer reflection from a denser medium of surface AB
We known that for constructive interference Path difference`=mlamda` where m is 1,2,...
`:'` `_mnxx2t+(lamda)/(2)=mlamdaimplies2` `_mnt=(m-(1)/(2))lamda`
when `m=1`, t=(lamda)/(4_mn)=(648)/(4xx1.8)=90`nm.`