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  • In the situation illustrated below, the object is set up 0.5 m away from a converging mirror. If the focal length of the lens is 0.2 m, determine a) the location of the real image, and b) the magni

In the situation illustrated below, the object is set up 0.5 m away from a converging mirror. If the focal length of the lens is 0.2 m, determine a) the location of the real image, and b) the magnification of the image.

Option: 1

v = 0.33 m \ \ , \ m = 0.33


Option: 2

v = 0.66 m \ \ , \ m = 0.33


Option: 3

v = 0.66 m \ \ , \ m = 0.66


Option: 4

v = 0.33 m \ \ , \ m = 0.66


Answers (1)

\begin{array}{l}{\text { In order to determine the location of the image, we'll use the Mirror Formula. }} \\ \\ {\qquad \frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \\ \\ {u=-0.5 m, f=-0.2 m} \\ \\ {\frac{1}{v}=-\frac{1}{0.2}+\frac{1}{0.5}} \\ \\ {v=-0.33 m}\end{array}

Here “–” sign shows that the image is real.

\begin{array}{l}{\text { Now that we have the location of the image, we can find the magnification. }} \\ {\qquad \begin{aligned} m &=-\frac{v}{u} \\ &=-\frac{(-0.33)}{(-0.5)} \\ &=-0.66 \end{aligned}}\end{array}

As |m|<1, the size of the image is smaller than the object and the negative sign shows that the image is real and inverted.

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Kshitij

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