Solveeit Logo

Question

Physics Question on Ray optics and optical instruments

A short linear object, of length ll, lies along the axis of a concave mirror, of focal length ff, at a distance dd from the pole of the mirror. The size of the image is then (nearly)

A

lfdf\frac{lf}{d-f}

B

dflf\frac{d-f}{lf}

C

lf2(df)2l\frac{f^{ 2}}{\left(d-f\right)^{2}}

D

(df)2f2l\frac{\left(d-f\right)^{2}}{f^{ 2}}l

Answer

lf2(df)2l\frac{f^{ 2}}{\left(d-f\right)^{2}}

Explanation

Solution

Use 1υ+1u=1f\frac{1}{\upsilon}+\frac{1}{u}=\frac{1}{f} Find υ1,\upsilon_{1}, when u=du = d and υ2,\upsilon_{2}, when u=d+lu = d + l Size of image =υ1υ2=l(fdf)2=\upsilon_{1}-\upsilon_{2}=l\left(\frac{f}{d-f}\right)^{2}