JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 10)
A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of
curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :
$${R \over 3}$$
$${{3R} \over 2}$$
$${R \over 2}$$
2R
Explanation
Assume refractive index = $$\mu $$$$l$$
P = $$\left( {{\mu _l} - 1} \right)\left( {{2 \over R}} \right)$$ .....(1)
$${3 \over 2}P = \left( {{\mu _l} - 1} \right)\left( {{1 \over {{R_1}}}} \right)$$ ......(2)
from (1)/(2)
$${P \over {{3 \over 2}P}} = {{\left( {{2 \over R}} \right)} \over {\left( {{1 \over {{R_1}}}} \right)}}$$
$$ \Rightarrow $$ R1 = $${R \over 3}$$
P = $$\left( {{\mu _l} - 1} \right)\left( {{2 \over R}} \right)$$ .....(1)
$${3 \over 2}P = \left( {{\mu _l} - 1} \right)\left( {{1 \over {{R_1}}}} \right)$$ ......(2)
from (1)/(2)
$${P \over {{3 \over 2}P}} = {{\left( {{2 \over R}} \right)} \over {\left( {{1 \over {{R_1}}}} \right)}}$$
$$ \Rightarrow $$ R1 = $${R \over 3}$$
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