Question

Focal length of an equi convex is numerically equal to its radius. The speed of the light in lens material is nearly equal to ​

Answer 1

Given:

Focal length of an equi convex is numerically equal to its radius.

To find:

Approximate velocity of light in the lens material ?

Calculation:

General expression for lens maker's formula for a convex lens is :

$\boxed{ \bold{ \dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{1}{R_{1}} - \dfrac{1}{R_{2}} \bigg \}}}$

Now, since the lens is equiconvex, R1 and R2 is equal.

Putting R1 = R2 = R (let) following sign convention:

$\implies \sf\dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{1}{R_{1}} - \dfrac{1}{R_{2}} \bigg \}$

$\implies \sf\dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{1}{R} - \bigg(\dfrac{1}{ - R} \bigg) \bigg \}$

$\implies \sf\dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{1}{R} + \dfrac{1}{ R} \bigg \}$

$\implies \sf\dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{2}{R} \bigg \}$

Now, focal length is numerical equal to the radius of curvature:

$\implies \sf\dfrac{1}{f} = (\mu - 1) \bigg \{ \dfrac{2}{f} \bigg \}$

$\implies \sf\dfrac{1}{2} = \mu - 1$

$\implies \sf \mu = 1 + \dfrac{1}{2}$

$\implies \sf \mu = \dfrac{3}{2}$

Now, velocity of light in that material:

$\sf\therefore \: v = \dfrac{velocity \: in \: vacuum}{ \mu}$

$\sf\implies\: v = \dfrac{3 \times {10}^{8} }{ (\frac{3}{2} )}$

$\sf\implies\: v = 2 \times {10}^{8} \: m {s}^{ - 1}$

So, velocity of light in the lens is 2 × 10⁸ m/s.