Question

A lens made of glass whose index of refraction is $1.60$ has a focal length of $+20cm$ in air. Its focal length in water, whose refractive index is $1.33$ , will be

• A. three times longer than in air
• B. two times longer than in air
• C. same as in air
• D. None of the above

Answer 1

Answer: (A)

Given that,
the refractive index of the lens wrt air, ${}_a{\mu }_g=1.60$
and the refractive index of water wrt air
${}_a{\mu }_w=1.33$
the focal length of the lens in air, $f=20cm$.
We know that for a lens
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
When the lens is in the air
$\frac{1}{20}=\left({}_a{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{20}=\left(1.60-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{20}=0.60\times \left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots \left(i\right)$
When the lens is in the water
$\frac{1}{f^{\prime }}=\left({}_w{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{f^{\prime }}=\left(\frac{{}_a{\mu }_g}{{}_a{\mu }_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{f^{\prime }}=\left(\frac{{}_a{\mu }_g{-}_a{\mu }_w}{{}_a{\mu }_w}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \frac{1}{f^{\prime }}=\left(\frac{1.60-1.33}{1.33}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{f^{\prime }}=\frac{27}{133}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\dots \left(ii\right)$
On dividing Eq. $\left(i\right)$ by Eq. $\left(ii\right)$, we get
$\frac{f^{\prime }}{20}=\frac{0.60\times 133}{27}$
or $f^{\prime }=20\times 2.95cm\approx 60cm$
Hence, its focal length is three times longer than in air