Optics Question 788

Question: When an object is placed at a distance of 25 cm from a mirror, the magnification is $ m _{1} $ . The object is moved 15cm further away with respect to the earlier position, and the magnification becomes If $ m _{1}/m _{2}=4, $ the focal length of the mirror is:

Options:

A) 10 cm

B) 30 cm

C) 15 cm

D) 20 cm

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $ m=-\frac{v}{u}-( \frac{f}{u-f} ) $

$ \text{Now }m _{1}=-( \frac{f}{25-f} ) $ –(i)

$ \text{and }m _{2}=-( \frac{f}{40-f} ) $ –(ii)

$ \therefore \frac{m _{1}}{m _{2}}=\frac{40-f}{25-f}\text{ or }4=\frac{40-f}{25-f}\text{ or f=20}cm\text{.} $



NCERT Chapter Video Solution

Dual Pane