Question

The diameter of the plano-convex lens is $6cm$ and the thickness at the center is $3mm$. If the speed of light in the material of the lens is $2\times {10}^{8}m/s$, the focal length of the lens is:

• A. $20cm$
• B. $30cm$
• C. $10cm$
• D. $15cm$

Answer 1

The correct option is B

$30cm$

Step 1: Given Data

The diameter of the plano-convex lens is $6cm$.

The thickness at the center = y = $3mm$= $\frac{3}{10}cm$

The speed of light in the material of the lens is $2\times {10}^{8}m/s$.

Let the focal length be $f$.

Let the refractive index be $\mu$.

Let the radius of curvature be $R$.

Speed of light in vacuum $c=3\times {10}^{8}m/s$

Here r =$\frac{diameteroftheplano-convexlens}{2}=\frac{6cm}{2}=3cm$

Step 2: Formula Used

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}\right)$

Step 3: Calculations

The lens is as shown in the figure.
To determine the radius of curvature $R$ of the curved surface, we can write

$R^2=r^2+{\left(R-y\right)}^2$
$R^2=3^2+{\left(R-\frac{3}{10}\right)}^2$

Solving this, we get
$R=15.15cm$
Speed of light in any medium is $\frac{c}{\mu }$ where here $c=3\times {10}^{8}m/s$ and $\mu$ is the refractive index.
For the lens, $\frac{c}{\mu }$$=2\times {10}^{8}m/s$

Putting these values in the lens makers' formula for the plano-convex lens, we have,

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R}\right)=\left(1.5-1\right)\left(\frac{1}{15.15}\right)c{m}^{-1}$

Solving this, we get $f\approx 30cm$.

Hence, the correct option is B.