Question

A concave lens forms the image of an object such that the distance between the object and image is $10cm$. If magnification of the image is $\frac{1}{4}$. The focal length of the lens is :

• A. $6cm$
• B. $\frac{20}{3}cm$
• C. $\frac{10}{7}cm$
• D. $\frac{-40}{9}$

Answer 1

The correct option is D $\frac{-40}{9}$

Given,

$image=10cm,magnification=\frac{1}{4}$

Since the object and the image in the same direction,

$u-v=10cm\Rightarrow v=u-10$

Since the magnification is positive and less than its virtual image,

Such that, $m=\frac{v}{u}=\frac{u-10}{u}=\frac{1}{4}\Rightarrow 4(u-10)=u\Rightarrow u=\frac{-40}{3}$

So $v=\frac{10}{3}$

Now the focal length by the lens formula,

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

$\frac{1}{f}=\frac{-3}{10}+\frac{3}{40}=\frac{-9}{40}\Rightarrow f=\frac{-40}{9}$