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{.} $
