JEE MAIN - Physics (2004 - No. 37)
A plano convex lens of refractive index $$1.5$$ and radius of curvature $$30$$ $$cm$$. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object
$$60$$ $$cm$$
$$30$$ $$cm$$
$$20$$ $$cm$$
$$80$$ $$cm$$
Selitys
KEY CONCEPT : The focal length $$\left( F \right)$$ of the final mirror
is $${1 \over F} = {2 \over {f\ell }} + {1 \over {{f_m}}}$$
Here $${1 \over {{f_\ell }}} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ = \left( {1.5 - 1} \right)\left[ {{1 \over \alpha } - {1 \over { - 30}}} \right] = {1 \over {60}}$$
$$\therefore$$ $${1 \over F} = 2 \times {1 \over {60}} + {1 \over {30/2}} = {1 \over {10}}$$
$$\therefore$$ $$F=10cm$$
The combination acts as a converging mirror. For the object to be of the same size of mirror,
$$u = 2F = 20cm$$
is $${1 \over F} = {2 \over {f\ell }} + {1 \over {{f_m}}}$$
Here $${1 \over {{f_\ell }}} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ = \left( {1.5 - 1} \right)\left[ {{1 \over \alpha } - {1 \over { - 30}}} \right] = {1 \over {60}}$$
$$\therefore$$ $${1 \over F} = 2 \times {1 \over {60}} + {1 \over {30/2}} = {1 \over {10}}$$
$$\therefore$$ $$F=10cm$$
The combination acts as a converging mirror. For the object to be of the same size of mirror,
$$u = 2F = 20cm$$