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Q.
The focal length of a convex lens dipped in a liquid of same refractive index as that of the lens, will:
NTA AbhyasNTA Abhyas 2020
Solution:
Lens maker's formula is given by,
$\frac{1}{f}=\left(\right.\left(\mu \right)_{2}-\left(\mu \right)_{1}\left.\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$ , where $\mu _{2}$ is the refractive index of the material of lens and $\mu _{1}$ is the refractive index of the medium in which lens is being kept.
If $\mu _{1}=\mu _{2}$ . Then,
$\frac{1}{f}=\left(\right.\left(\mu \right)_{2}-\left(\mu \right)_{2}\left.\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)\Rightarrow \frac{1}{f}=0\Rightarrow f=\infty $
Then lens behaves like a plane surface with focal length infinity.