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An object of length 2.5 \mathrm{~cm} is placed at a 1.5 \mathrm{f} from a concave mirror where \mathrm{f} is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. The length of the image is:

Option: 1

2 \mathrm{~cm}


Option: 2

3 \mathrm{~cm}


Option: 3

5 \mathrm{~cm}


Option: 4

7 \mathrm{~cm}


Answers (1)

best_answer

The focal length \mathrm{F=-f}

\mathrm{ \text { and } \quad u=-1.5 \mathrm{f} \text {, we have, } }

\mathrm{ \frac{1}{u}+\frac{1}{v}=\frac{1}{F} \quad \text { or } \quad \frac{1}{-1.5 f}+\frac{1}{v}=-\frac{1}{f}}
\mathrm{ \frac{1}{v}=\frac{1}{1.5 f}-\frac{1}{f}=-\frac{1}{3 f}, \quad v=-3 f}
\mathrm{ \text { Now, } \quad m=-v / u=-\frac{3 f}{1.5 f}=-2}
\mathrm{ \text { or } \frac{h_{2}}{h_{1}}=-2 \quad \text { or } \quad h_{2}=-2 h_{1}=-5 \mathrm{~cm}}


The image is 5 \mathrm{~cm} long. The minus sign shows that it is inverted.

Posted by

vinayak

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