Question

As shown in figure, a curved surface of radius R = 20 cm seperate two medium of refractive index ${\mu }_1=1$ and ${\mu }_2=3/2$. A point at 50 cm from curved surface starts moving towards surface at 4 m/s, then

• A. virtual image of object is formed
• B. real image of object is formed
• C. image is formed at 200 cm from curved surface
• D. speed of image is 96 m/s

Answer 1

Answer: (B), (D)

The correct options are
B real image of object is formed
D speed of image is 96 m/s

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\Rightarrow \frac{3}{2v}+\frac{1}{50}=\frac{1}{40}$
$v=300cm$

We know that,

$-\frac{{\mu }_2}{v^2}\frac{dv}{dt}+\frac{{\mu }_1}{u^2}\frac{du}{dt}=0\Rightarrow \frac{dv}{dt}=\frac{{\mu }_1}{{\mu }_2}{\left(\frac{v}{u}\right)}^2\frac{du}{dt}\frac{dv}{dt}=\frac{2}{3}{\left(\frac{300}{-50}\right)}^2\times 4=96m/s$