Question

A lens made of glass whose index of refraction is $1.60$ has a focal length of $+20\, cm$ 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, $f = 20 \,cm$.
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 $\frac{1}{20}=0.60\times\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots\left(i\right)$
When the lens is in the water
$\frac{1}{f'}=\left(_{w}\mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
or $\frac{1}{f'}=\left(\frac{_{a}\mu_{g}}{_{a}\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
or $\frac{1}{f'} =\left(\frac{_{a}\mu_{g}-_{a}\mu_{w}}{_{a}\mu_{w}}\right) \left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$\therefore \frac{1}{f'}=\left(\frac{1.60-1.33}{1.33}\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
or $\frac{1}{f'}=\frac{27}{133} \left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \dots\left(ii\right)$
On dividing Eq. $\left(i\right)$ by Eq. $\left(ii\right)$, we get
$\frac{f'}{20}=\frac{0.60\times133}{27}$
or $f' =20\times2.95\,cm \approx60\, cm$
Hence, its focal length is three times longer than in air