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A real image of half the size is obtained in a concave spherical mirror with a radius of curvature of \mathrm{40\ cm}. The distance of object and its image will be:

Option: 1

\mathrm{30\ cm\ and\ 60\ cm}


Option: 2

\mathrm{60\ cm\ and\ 30\ cm}


Option: 3

\mathrm{15\ cm\ and\ 30\ cm}


Option: 4

\mathrm{30\ cm\ and\ 15\ cm}


Answers (1)

best_answer

Since the image is real
\begin{aligned} & \mathrm{\therefore \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{1}{2} \text { or } \mathrm{u}=2 \mathrm{v} }\\ & \therefore \quad \mathrm{R}=40 \mathrm{~cm} \Rightarrow \mathrm{f}=-20 \mathrm{~cm} \end{aligned}
\text { As } \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \ \ \ \quad \therefore \frac{1}{\mathrm{v}}+\frac{1}{2 \mathrm{v}}=\frac{1}{-20}
\text { or } \quad\mathrm{ \frac{3}{2 v}}=-\frac{1}{20}\ \ \quad \text { or } \mathrm{v}=-30 \mathrm{~cm}\ \ \quad \text { or } \quad \mathrm{u}=2 \mathrm{v}=-60 \mathrm{~cm}

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shivangi.bhatnagar

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