Question 98: Ray Optics

Question

A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water.
(Take refractive index of water as 1.28)

Accepted Answer

To find the focal length of a lens with a refractive index of 1.6 when placed in water, given that its focal length in air is 12 cm and the refractive index of water is 1.28, we can use the lens maker's formula:
$ \frac{1}{f} = \left(\frac{\mu_L}{\mu_m} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $
When the lens is in air, the refractive index of the medium, $\mu_m$, is 1. Therefore, for the lens in air:
$ \frac{1}{12} = (1.6 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $
Simplifying gives:
$ \frac{1}{12} = \frac{6}{10} \left(\frac{1}{R_1} - \frac{1}{R_2}ight) $
Solving for $\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$:
$ \frac{1}{R_1} - \frac{1}{R_2} = \frac{10}{72} $
Now, when the lens is placed in water, $\mu_m = 1.28$. Substituting into the equation gives:
$ \frac{1}{f} = \left(\frac{1.6}{1.28} - 1ight) \left(\frac{10}{72}ight) $
Calculating each part:
$ \frac{1}{f} = \frac{32}{128} 	imes \frac{10}{72} $
$ \frac{1}{f} = \frac{1}{4} 	imes \frac{10}{72} $
Hence, solving for $f$:
$ f = 28.8 \, 	ext{cm} = 288 \, 	ext{mm} $
Thus, the focal length of the lens in water is 288 mm.

Answer

355 mm

Answer

655 mm

Answer

288 mm

Answer

555 mm

Ray Optics

A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

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