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

Physics · Geometrical Optics

JEE Main 2025 — 4 April, Morning Shift — Question 66

Distance between object and its image (magnified by $\left.-\frac{1}{3}\right)$ is 30 cm .

The focal length of the mirror used is $\left(\frac{x}{4}\right) \mathrm{cm}$ , where magnitude of value of

$x$ is $\qquad$

Numerical answer — enter this value.

$m=\frac{f}{f-\mu}=-\frac{1}{3}$

$4 f=\mu$

$\frac{1}{v}+\frac{1}{4 f}=\frac{1}{f}$

$v=\frac{4 f}{3}$

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

$f=\frac{90}{8}=\frac{45}{4}$

$x=45$

Answer key and solution verified before publishing.

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