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Question

Physics Question on Ray optics and optical instruments

A short linear object of length bb lies along the axis of a concave mirror of focal length ff at a distance uu from the pole of the mirror, what is the size of image?

A

(fuf)b\left( \frac{f}{u-f} \right)b

B

(fuf)2b{{\left( \frac{f}{u-f} \right)}^{2}}b

C

(fuf)b2\left( \frac{f}{u-f} \right){{b}^{2}}

D

(fuf)\left( \frac{f}{u-f} \right)

Answer

(fuf)2b{{\left( \frac{f}{u-f} \right)}^{2}}b

Explanation

Solution

Using the relation for the focal length of concave mirror 1f=1v+1u\frac{1}{f}=\frac{1}{v}+\frac{1}{u} ... (i) Differentiating E (i), we obtain 0=1v2dv1u2du0=-\frac{1}{v^{2}} d v-\frac{1}{u^{2}} d u So, dv=v2u2×bd v=-\frac{v^{2}}{u^{2}} \times b ...(ii) (Here, du=bdu=b ) From E (i) 1v=1f1u=uffv\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f v} or uv=uff\frac{u}{v}=\frac{u-f}{f} vu=fufˉ\frac{v}{u}=\frac{f}{u-\bar{f}} ...(iii) Now, from Eqs. (ii) and (iii), we get dv=(fuf)2bd v=-\left(\frac{f}{u-f}\right)^{2} b Therefore, size of image is =(fuf)2b=\left(\frac{f}{u-f}\right)^{2} b