JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 22)
Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm . The focal length of the mirror used is $\left(\frac{x}{4}\right) \mathrm{cm}$,
where magnitude of value of $x$ is _________.
Answer
45
Explanation
$$\begin{aligned} &\begin{aligned} & M=-\frac{1}{3} \\ & -\frac{-V}{-U}=\frac{-1}{3} \Rightarrow V=\frac{U}{3} \end{aligned}\\ &\text { Distance b/w object and image : } \end{aligned}$$
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$$\begin{aligned} & U-V=30 \\ & U-\frac{\mu}{3}=30 \\ & \Rightarrow U=45 \quad V=15 \\ & \frac{1}{f}=\frac{1}{V}+\frac{1}{U}=-\frac{1}{15}-\frac{1}{45} \\ & \Rightarrow F=\frac{45}{4} \\ & x=45 \end{aligned}$$
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