JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 6)
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :
$${{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
$${{2{d_1}{d_2}} \over {{d_1} + {d_2}}}$$
$${{{d_1}{d_2}} \over {{d_1} + {d_2}}}$$
$${{{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
Explanation
Using Newton's formula
$$(f + {d_1})(f - {d_2}) = {f^2}$$
$${f^2} + f{d_1} - f{d_2} - {d_1}{d_2} = {f^2}$$
$$f = {{{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
$$\therefore$$ $$R = {{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
$$(f + {d_1})(f - {d_2}) = {f^2}$$
$${f^2} + f{d_1} - f{d_2} - {d_1}{d_2} = {f^2}$$
$$f = {{{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
$$\therefore$$ $$R = {{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$$
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