Question

A lens made of glass of refractive index 1.5 has focal length 25 cm its focal length in water will be
R I in water 4/3

Answer 1

Answer:

Focal length in water = (In air)/ ur

$ur = \frac{ \frac{4}{3} }{ \frac{3}{2} } = 8 \\ focal \: length \: in \: water = \frac{25}{8}$

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Answer 2

The focal length of the lens in the water is 97.79 cm.

Given,

Refractive index of glass = 1.5

Focal length in air = 25 cm

Refractive index in water = 4/3 = 1.33

To Find,

The focal length of the lens in the water.

Solution,

We know that 1/f = (μ - 1)(1/R1 - 1/R2)

⇒ 1/f = (μg/μa - 1)(1/R1 - 1/R2)

⇒ 1/25 = (1.5/1 - 1)(1/R1 - 1/R2)

⇒ 1/25 = (0.5)(1/R1 - 1/R2)

⇒ (1/R1 - 1/R2) = 1/25x0.5                            ---------------------------------------(1)

For water,

1/fw = (μg/μw - 1)(1/R1 - 1/R2)

On putting the value of (1/R1 - 1/R2) from (1), we get,

1/fw = (μg/μw - 1)(1/25x0.5)

⇒ 1/fw = (1.5/1.33 - 1)(1/12.5)

⇒ 1/fw = (0.17/1.33)(1/12.5)

⇒ 1/fw = 1/97.79

⇒ fw = 97.79 cm

Hence, the focal length of the lens in the water is 97.79 cm.

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