Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose.(i) State the nature of the lens and the reason for its use.(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?(iii) If the focal length of this lens is 10 cm and the lens is held at a distance of 5 cm from the palm, use the lens formula to find the position and size of the image.

(i) The nature of the lens will be convex lens and the reason for its use is, it gives a magnified image of the lines on the palm.

(ii) The palmist should place/hold the lens between $F'$ and $2F'$ of the lens or at $F'$ of the lens so as to have a real and magnified image of an object.

(iii) Given:

Focal length, $f$ = $+$10 cm   (focal length of the convex lens is always taken positive)

Object distance, $u$ = $-$5 cm  (object distance is always taken negative as it placed on the left side of the lens)

To find: Position or distance of the image $v$, and size of the image $h'$.

Solution:

From the lens formula, we know that-

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values we get-

$\frac {1}{v}-\frac {1}{(-5)}=\frac {1}{10}$

$\frac {1}{v}+\frac {1}{5}=\frac {1}{10}$

$\frac {1}{v}=\frac {1}{10}-\frac {1}{5}$

$\frac {1}{v}=\frac {1-2}{10}$

$\frac {1}{v}=\frac {-1}{10}$

$v=−10cm$

Thus, the position or distance of the image, $v$ is 10cm from the lens, and the negative sign implies that it forms on the left side of the lens.

Now, from the magnification formula, we know that-

$m=\frac {v}{u}$

$m=\frac {-10}{-5}$

$m=+2$

Thus, the size of the image, $h'$ is 2 times the size of the object.

Tutorialspoint

Updated on: 10-Oct-2022

8K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started