Question

The radius of curvature of curved surface of a plano-convex lens is $15\text{cm}$. The refractive index of the lens material is $1.5$. The focal length of plano-convex lens is (surroundings is air)

• A. $20\text{cm}$
• B. $-40\text{cm}$
• C. $30\text{cm}$
• D. $25\text{cm}$

Answer 1

The correct option is C $30\text{cm}$





For a plane surface, $R_1=\infty$

For curved surface, $R_2=-15\text{cm}$

Also, ${\mu }_1=1,{\mu }_2=1.5$

Applying lens maker's formula:

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow \frac{1}{f}=(1.5-1)(\frac{1}{\infty }-\frac{1}{-15})$

$\Rightarrow \frac{1}{f}=\left(0.5\right)\left(\frac{1}{15}\right)=\frac{1}{30}$

$\therefore f=+30\text{cm}$

Why this question? Tip: In such questions, first of all direction of incidence must be set to avoid any confusions. Thereafter, apply the lens maker's formula.