Question

The focal length of thin
convex lens is 20 cm. When
an object is moved from a
distance of 25 cm infront of
it to 50 cm, the magnification
of its image changes from
m25 to m50. The ratio of m25/
m50 is​

Answer 1

Given that,

Focal length = 20 cm

Object distance $u_{25}= 25\ cm$

Object distance $u_{50}= 50\ cm$

When u = -25 cm

We need to calculate the image distance

Using formula of lens

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

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-25}$

$\dfrac{1}{v}=\dfrac{1}{100}\ cm$

$v=100\ cm$

We need to calculate the magnification

Using formula of magnification

$m_{25}=\dfrac{-v}{u_{25}}$

Put the value into the formula

$m_{25}=\dfrac{-100}{-25}$

$m_{25}=4$.....(I)

When u = -50 cm

We need to calculate the image distance

Using formula of lens

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

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-50}$

$\dfrac{1}{v}=\dfrac{3}{100}\ cm$

$v=\dfrac{100}{3}\ cm$

We need to calculate the magnification

Using formula of magnification

$m_{50}=\dfrac{-v}{u_{50}}$

Put the value into the formula

$m_{50}=\dfrac{-100}{-50\times3}$

$m_{50}=\dfrac{2}{3}$.....(II)

We need to calculate the ratio of $m_{25}$ and $m_{50}$

Using equation (I) and (II)

$\dfrac{m_{25}}{m_{50}}=\dfrac{4\times3}{2}$

$\dfrac{m_{25}}{m_{50}}=6$

Hence, The ratio of $m_{25}$ and $m_{50}$ is 6.