Question

A thin symmetrical double convex lens of refractive index ${\mu }_2=1.5$ is placed between a medium of refractive index ${\mu }_1=1.4$ to the left and another medium of refractive index ${\mu }_3=1.6$ to the right. Then, the system behaves as.

• A. A convex lens
• B. A concave lens
• C. A glass plate
• D. A convo concave lens

Answer 1

The correct option is C A glass plate

For first refracting Surface,

For parallel beam of light, $-\frac{{\mu }_1}{u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-1}{R}$

$-\frac{1.4}{-\infty }+\frac{1.5}{v}=\frac{1.5-1.4}{R}$

So V = 15R

For Second refracting surface,

$-\frac{{\mu }_2}{1.5R}+\frac{{\mu }_3}{v^{\prime }}=\frac{{\mu }_3-{\mu }_2}{R}$

$-\frac{1.5}{15R}+\frac{1.6}{v^{\prime }}=\frac{1.6-1.5}{R}$

$\therefore v^{\prime }=\infty$

Hence, the combination behaves as glass plate.