Question

A rod of length $10\, cm$ lies along the principal axis of a concave mirror of focal length $10\, cm$ in such a way that its end closer to the pole is $20\, cm$ away from the mirror. The length of the image is

• A. 10 cm
• B. 15 cm
• C. 2.5 cm
• D. 5 cm

Answer 1

The Correct Option is D

Here,/= -10 cm
For end A ,$ u_A $ = - 20 cm
Image position of end A.
$\frac{ 1}{{\upsilon}_A}+ \frac{1}{u_A} = \frac{1}{f}$
$\frac{1}{{\upsilon}_A}+ \frac{1}{(-20)} = \frac{ 1}{(-10)} \, \, or = \frac{1}{{\upsilon}_A} = \frac{ 1}{-10 } +\frac{1}{20} = -\frac{1}{20}$
${\upsilon}_A = -20 cm $
For end B, $u_B $= - 30 cm
Image position of end B,
$\frac{ 1}{{\upsilon}_B}+ \frac{1}{u_B} = \frac{1}{f}$
$\frac{1}{{\upsilon}_B}+ \frac{1}{(-30)} = \frac{ 1}{(-10)} \, \, or = \frac{1}{{\upsilon}_B} = \frac{ 1}{-10 } +\frac{1}{30} = -\frac{2}{30}$
${\upsilon}_B= -15 cm $
Length of the image
$=| {\upsilon}_A|-| {\upsilon}_B| = 20cm -15 cm = 5cm $