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Physics Question on Ray optics and optical instruments

A rod of length 10 cm10\, cm lies along the principal axis of a concave mirror of focal length 10 cm10\, cm in such a way that its end closer to the pole is 20 cm20\, 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

5 cm

Explanation

Solution

Here,/= -10 cm
For end A ,uAu_A = - 20 cm
Image position of end A.
1υA+1uA=1f\frac{ 1}{{\upsilon}_A}+ \frac{1}{u_A} = \frac{1}{f}
1υA+1(−20)=1(−10)  or=1υA=1−10+120=−120\frac{1}{{\upsilon}_A}+ \frac{1}{(-20)} = \frac{ 1}{(-10)} \, \, or = \frac{1}{{\upsilon}_A} = \frac{ 1}{-10 } +\frac{1}{20} = -\frac{1}{20}
υA=−20cm{\upsilon}_A = -20 cm
For end B, uBu_B = - 30 cm
Image position of end B,
1υB+1uB=1f\frac{ 1}{{\upsilon}_B}+ \frac{1}{u_B} = \frac{1}{f}
1υB+1(−30)=1(−10)  or=1υB=1−10+130=−230\frac{1}{{\upsilon}_B}+ \frac{1}{(-30)} = \frac{ 1}{(-10)} \, \, or = \frac{1}{{\upsilon}_B} = \frac{ 1}{-10 } +\frac{1}{30} = -\frac{2}{30}
υB=−15cm{\upsilon}_B= -15 cm
Length of the image
=∣υA∣−∣υB∣=20cm−15cm=5cm=| {\upsilon}_A|-| {\upsilon}_B| = 20cm -15 cm = 5cm