Question

Find the size of the image formed in the situation shown in figure.

• A. 0.8 cm, inverted
• B. 0.6 cm, inverted
• C. 0.6 cm, erect
• D. 0.4 cm, erect

Answer 1

The correct option is C

0.6 cm, erect

Here
$u=-40\text{cm},R=-20\text{cm},$
${\mu }_1=1,{\mu }_2=\mathrm{1.33.}$

We have,
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\frac{1.33}{v}-\frac{1}{-40}=\frac{1.33-1}{-20}$
$\frac{1.33}{v}=-\frac{1}{40}-\frac{0.33}{20}$
$v=-32\text{cm}.$
The magnification is,
$m=\frac{h_2}{h_1}=\frac{{\mu }_{1}v}{{\mu }_{2}u}$
$\frac{h_2}{1.0}=\frac{-32}{1.33\times (-40)}$
$h_2=+0.6$
The image is erect as the magnification is positive.