Discussion Forum

A thin convex lens made from crown glass (μ =3/2) has focal length f. When it is measured in two different liquids having refractive index 4/3 and 5/3. It has the focal length f₁  and f₂ respectively. The correct relation between  the focal length is 

Answer:

Explanation:

It is based on lens makers formula and its magnification

 i.e,   $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

According to lens Maker's formula,when the lens in the air.

$\frac{1}{f}=\left( \frac{3}{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\frac{1}{f}=\frac{1}{2x}$

$\Rightarrow$            $ f=2x$

    $\left(\frac{1}{x}=\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

In case of liquid , where  refractive index is  $\frac{4}{3}$ and   $\frac{5}{3}$, we get

focal length in first liquid 

$\frac{1}{f_{1}}=\left(\frac{\mu_{s}}{\mu_{l1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\Rightarrow$          $\frac{1}{f_{1}}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)\frac{1}{x}$

$\Rightarrow$            f₁  is positive

$\frac{1}{f_{1}}=\frac{1}{8x}=\frac{1}{4(2x)}=\frac{1}{4f}$

$\Rightarrow$             f₁=4f

  focal length in second liquid

            $\frac{1}{f_{2}}=\left(\frac{\mu_{s}}{\mu_{l1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

$\Rightarrow$          $\frac{1}{f_{2}}=\left(\frac{\frac{3}{2}}{\frac{5}{3}}-1\right)\frac{1}{x}$

$\Rightarrow$            f₂ is negative