Question 4: Ray Optics

Question

A lens of refractive index $$\mu$$ is put in a liquid of refractive index $$\mu^{\prime}$$. If the focal length of lens in air is $$f$$, then its focal length in liquid will be

Accepted Answer

According to question,
Here, $$\mu=$$ refractive index of lens
$$\mu^{\prime}=$$ refractive index of liquid focal length of lens in air $$=f$$
We have lens maker's formula
$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$$
When lens is dipped in water,
$$\frac{1}{f^{\prime}}=\left(\frac{\mu_g}{\mu_l}-1ight)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$$
or $$\left(\frac{\mu_g-\mu_l}{\mu_l}ight)\left(\frac{1}{R_1}-\frac{1}{R_2}ight)$$
$$	herefore \quad \frac{f^{\prime}}{f}=\frac{(\mu-1) \mu^{\prime}}{\left(\mu-\mu^{\prime}ight)}=\frac{-(\mu-1) \mu^{\prime}}{\left(\mu^{\prime}-\muight)} \quad\left(\because \mu_l=\mu^{\prime}ight)$$
or $$\quad f^{\prime}=\frac{-f \mu^{\prime}(\mu-1)}{\left(\mu^{\prime}-\muight)}$$

Answer

$$\frac{-f \mu^{\prime}(\mu-1)}{\left(\mu^{\prime}-\mu\right)}$$

Answer

$$\frac{-f\left(\mu^{\prime}-\mu\right)}{\mu^{\prime}(\mu-1)}$$

Answer

$$-\frac{\mu^{\prime}(\mu-1)}{f\left(\mu^{\prime}-\mu\right)}$$

Answer

$$\frac{f \mu^{\prime} \mu}{\left(\mu-\mu^{\prime}\right)}$$

Ray Optics

A lens of refractive index $\mu is put in a liquid of refractive index \mu^{\prime}. If the focal length of lens in air is f$, then its focal length in liquid will be

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