Distance between object and its image (magnified by – 1/3 ) is 30 cm. The focal length of the mirror used is (x/4)cm, ​

Question

Distance between object and its image (magnified by – \(\frac{1}{3}\)) is 30 cm. The focal length of the mirror used is \(\left(\frac{x}{4}\right)\)cm,

where magnitude of value of x is _____.

Answer 1

Answer is : (45)

\(m = \frac{f}{f-\mu} = - \frac{1}{3}\)

\(4f = \mu\)

\(\frac{1}{v}+ \frac{1}{4f} = \frac{1}{f}\)

\(v = \frac{4f}{3}\)

\((u-v) = \frac{8f}{3} = 30\)

\(f = \frac{90}{8} = \frac{45}{4}\)

x = 45