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