Question

An object is placed at right angles to the principal axis of a concave mirror such that the image formed is half the size of the object. The object is now shifted to another position,displaced relative to the earlier position by 10 cm,and the size of the image now becomes one-fourth of the object. Find the initial distance of the object from the pole and also find the focal length of the mirror.

Answer 1

Answer:

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Answer 2

Answer:

The initial position of the pole from the mirror is 15cm.

The focal length of the mirror is equal to -5cm.

Explanation:

Let the distance of the object from the pole is x. Therefore $u=-x$.

Since the given mirror is concave and the image is diminished.

The image will be real and inverted so the magnification will be negative.

$m=-\frac{v}{u}=-\frac{1}{2}$  

$v=\frac{-u}{2}=\frac{-x}{2}$

Apply,  $\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$

      $-\frac{2}{x} -\frac{1}{x} =\frac{1}{f}$

       $-\frac{3}{x} =\frac{1}{f}$                                                       ....................(1)

On shifting the position of object by 10cm. The size of image becomes one fourth of object while earlier it was half. So the object is shifted 10cm away from the pole.

$u=-(x+10)$

$m=-\frac{v}{u}=-\frac{1}{4}$

$v=-\frac{u}{4}=-\frac{(x+10)}{4}$

Using, $\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$

$-\frac{4}{x+10} -\frac{1}{x+10} =\frac{1}{f}$

$-\frac{5}{x+10} =\frac{1}{f}$                                                                  ..............(2)

From eq.(1) and eq.(2),

$-\frac{3}{x} =-\frac{5}{x+10}$

∴  $3x+30=5x\\x=15cm$

Thus the object was initially 15cm away from the pole.

From eq.(2),

$-\frac{5}{x+10} =\frac{1}{f}$

$-\frac{5}{15+10} =\frac{1}{f}$

∴  $f=-5cm$