Question

A simple magnifying lens is used by watch repairer in such a way that an image is formed at 25 cm away from the eye. If he want to increase the magnification by 10 times, then the focal length of the lens should be
(a) 2.7 cm
(b) 2.7 mm
(c) 27 cm
(d) 10 cm​

Answer 1

Answer: 2.7 cm; option A

Explanation:

10= 1 + 25/fe

9= 25/ fe

fe= 25/9= 2.7cm

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Answer 2

A watch repairer uses a simple magnifying lens to create an image at a distance of $25 cm$from the eye. The focal length of the lens should be  "(a)$2.7 cm$" if he wants to raise the magnification by$10$ times.

Step by step explanation:

Given:

Distance of the image(v) : $25cm$

Magnification(m): $10$ times

To find:

Focal length of the lens

Solution:

Lens formula

                      $\frac1v-\frac1u=\frac1f$

where,

$u =$distance of object from pole,$v =$ distance of image from pole and$f =$focal length of the lens.

The object is positioned at a distance of $-u$cm and created at a distance of $v = -25 cm.$

magnification of  lens $m=\frac vu$

Substituting the values of m and v,

                         $10=\frac vu$

                     $10=\frac{-25}u$

                      $u=\frac{-25}{10}$

                      $u=-2.5 cm$

The focal length can be computed using the lens formula using the values of u and v.

                     $\frac1v-\frac1u=\frac1f$

         $\frac1{(-25)}-\frac1{(-2.5)}=\frac1f$

                  $\frac1{25}-\frac1{2.5}=\frac1f$

                             $\frac1f=\frac{10-1}{25}$

                            $f=\frac{25}9$

                           $f=2.7cm$

Focal length of the lens is$2.7cm$