Question 291: Ray Optics

Question

The radius of curvature of each surface of a convex lens having refractive index 1.8 is $$20 \mathrm{~cm}$$. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be $$x: 1$$. The value of $$x$$ is _________.

Accepted Answer

Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$
where $$f$$ is the focal length, $$\mu$$ is the refractive index of the lens material, and $$R_1$$ and $$R_2$$ are the radii of curvature of the lens surfaces.
Since the lens is convex, both surfaces have the same radius of curvature (positive), so $$R_1 = R_2 = 20 \mathrm{~cm}$$.
First, let's find the focal length of the lens in air:
$$\frac{1}{f_	ext{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}ight) = 0.8\left(\frac{1}{20}ight)$$
$$f_	ext{air} = \frac{1}{0.8\left(\frac{1}{20}ight)} = 25 \mathrm{~cm}$$
Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:
$$\mu_	ext{rel} = \frac{1.8}{1.5} = 1.2$$
$$\frac{1}{f_	ext{liquid}} = (1.2 - 1)\left(\frac{1}{20}ight)$$
$$f_	ext{liquid} = \frac{1}{0.2\left(\frac{1}{20}ight)} = 100 \mathrm{~cm}$$
The power of a lens is given by:
$$P = \frac{1}{f}$$
Now, we can find the ratio of the power of the lens in air to its power in the liquid:
$$\frac{P	ext{air}}{P
	ext{liquid}} = \frac{f	ext{liquid}}{f
	ext{air}} = \frac{100}{25} = 4$$
So, the ratio of the power of the lens in air to its power in the liquid is $$x:1$$, where $$x = 4$$.

Ray Optics

The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 \mathrm{~cm}. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be x: 1. The value of x$ is _________.

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