JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 16)
A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces.
_27th_June_Evening_Shift_en_16_1.png)
Power of $${L_1} = {P \over 2}$$
Power of $${L_2} = {P \over 2}$$
Power of $${L_3} = {P \over 2}$$
Power of L1 = P
Explanation
We know $$P = {1 \over f} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$${L_1}:{1 \over {{f_1}}} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) = {P_1} = (\mu - 1)\left( {{2 \over R}} \right) = P$$
$${L_2}:{1 \over {{f_2}}} = (\mu - 1)\left( {{1 \over {{R_1}}}} \right) = {P_2} = {{\mu - 1} \over R}$$
$${L_3}:{1 \over {{f_3}}} = (\mu - 1)\left( { - {1 \over {{R_2}}}} \right) = {P_3} = {{\mu - 1} \over R}$$
Comments (0)
