Object Pin Convex Lens Convex Mirror Image Pin
22.2 cm 32.2 cm 45.8 cm 71.2 cm

To find the focal length of a convex mirror, a student records the following data:
The focal length of the convex lens is $f_{1}$ and that of mirror is $f_{2}$ . Then taking index correction to be negligibly small, $f_{1}$ and $f_{2}$ are close to :

f_{1} = 7.8 c m f_{2} = 12.7 c m

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f_{1} = 12.7 c m f_{2} = 7.8 c m

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f_{1} = 7.8 c m f_{2} = 25.4 c m

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f_{1} = 15.6 c m f_{2} = 25.4 c m

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Solution

The correct option is A $f_{1} = 7.8 c m f_{2} = 12.7 c m$
Applying the lens equation, we get

$\frac{1}{v} + \frac{1}{10} = \frac{1}{f_{1}}$ ... (1)

For mirror,

$\frac{1}{25.4} - \frac{1}{45.8 - 32.2 - v} = \frac{1}{f_{2}}$

$\frac{1}{25.4} - \frac{1}{13.6 - v} = \frac{1}{f_{2}}$ .... (2)

Since there are 3 variables and only two equations (1) and (2) therefore only by substituting $f_{1}$ & $f_{2}$ from options correct values can be verified.
Eliminating $v$ from the equation (1), we get a relation between $f_{1}$ and $f_{2}$ . This relation is satisfied only by option A.

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A student has obtained an image of a distant on a screen to determine the focal length $F_{1}$ of the given lens. His teacher, after checking the image, gave him another lens of focal length $F_{2}$ and asked him to focus the same object on the same screen. The student found that to obtain a sharp image, he has to move the lens away from the screen. From this finding, we may conclude that both the lenses given to the student were: