Question

A single converging lens is held close to the eye for use as a mangnifying glass. For maximum magnifying power, the lens must be positioned, so that

• A. the image is at infinity.
• B. object is just beyond the principal focus of the lens.
• C. the image is at the principal focus of the lens.
• D. the image is at the near point of the eye.

Answer 1

The correct option is D the image is at the near point of the eye.


Magnifying power
= $\frac{\text{Angle subtended by the image on the eye when lens is used}}{\text{Angle subtended on naked eye when object is placed at near point}}$

Suppose object is placed at distance u${}_0$from lens, So that its image is formed at near point.

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

$\Rightarrow$ $\frac{1}{-D}$ $-$ $\frac{1}{-u_0}$ $=$ $\frac{1}{f}$

$\frac{1}{u_0}$$=$$\frac{1}{D}+\frac{1}{f}$
$\Rightarrow$ $\frac{D}{u_0}$ $=$ $1+\frac{D}{f}$

Angle subtended by image on lens (and hence, on eye) is

${\theta }^{{}^{\prime }}=\frac{h}{u_0}$

$\text{Magnifying power}=\frac{h/u_0}{h/D}=\frac{D}{u_0}=1+\frac{D}{f}$