JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 16)
A plano-convex lens (focal length f2, refractive index $$\mu $$2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index $$\mu $$1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :
f1 + f2
f1 $$-$$ f2
$${R \over {{\mu _2} - {\mu _1}}}$$
$${{2{f_1}{f_2}} \over {{f_1} + {f_2}}}$$
Explanation
_12th_January_Evening_Slot_en_16_1.png)
$${1 \over F} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{1 - {\mu _1}} \over R} + {{{\mu _2} - 1} \over R}$$
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