Question

Light is incident on a thin lens as shown in figure. The radius of curvature of both the surfaces is $R$. The focal length for this system will be

• A. $\frac{{\mu }_{3}R}{{\mu }_1}$
• B. $\frac{{\mu }_2}{R}{\mu }_2-{\mu }_1$
• C. $\frac{{\mu }_2-{\mu }_1}{R}$
• D. $\frac{{\mu }_{3}R}{{\mu }_3-{\mu }_1}$

Answer 1

The correct option is D $\frac{{\mu }_{3}R}{{\mu }_3-{\mu }_1}$

Since incident ray striking at first curved surface are parallel, it means object is at infinity.
$u=-\infty$
For refraction at first surface,

$\frac{{\mu }_2}{v_1}-\frac{{\mu }_1}{-\infty }=\frac{{\mu }_2-{\mu }_1}{+R}$

$\Rightarrow \frac{{\mu }_2}{v_1}=\frac{{\mu }_2-{\mu }_1}{R}.......\left(i\right)$

Now the image distance $v_1$ for refraction at first surface will serve as object distance for second curved surface.

$\frac{{\mu }_3}{v_2}-\frac{{\mu }_2}{v_1}=\frac{{\mu }_3-{\mu }_2}{+R}.....\left(ii\right)$

Focal length for system will be image distance for 2nd spherical surface.
sustituting in Eq. (ii) from Eq. (i),

$\frac{{\mu }_3}{v_2}-\frac{{\mu }_2-{\mu }_1}{R}=\frac{{\mu }_3-{\mu }_2}{R}$

$\Rightarrow \frac{{\mu }_3}{v_2}=\frac{({\mu }_3-{\mu }_2)+({\mu }_2-{\mu }_1)}{R}$

$\Rightarrow \frac{{\mu }_3}{v_2}=\frac{{\mu }_3-{\mu }_1}{R}$

$\Rightarrow v_2=\frac{{\mu }_{3}R}{{\mu }_3-{\mu }_1}$

$\therefore f=v_2=\frac{{\mu }_{3}R}{{\mu }_3-{\mu }_1}$

Why this question? It intends to showcase you "the application of refraction at a spherical surface, instead of lens maker's formula." Tip: The final image distance will give focal length of system.