JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 11)
If the distance between object and its two times magnified virtual image produced by a curved mirror is $$15 \mathrm{~cm}$$, the focal length of the mirror must be:
$$-10$$ cm
$$-12$$ cm
15 cm
10/3 cm
Explanation
_29th_January_Evening_Shift_en_11_1.png)
$$\begin{aligned} & \mathrm{m}=2=\frac{-v}{u} \\ & 2=\frac{-(15-u)}{-u} \\ & 2 u=15-u \\ & 3 u=15 \Rightarrow u=5 \mathrm{~cm} \\ & v=15-u=15-5=10 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \\ & f=-10 \mathrm{~cm} \end{aligned}$$
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