SATHEE: Optics 3 Question 30

Optics 3 Question 30

30. A thin lens of refractive index 1.5 has a focal length of $15 c m$ in air. When the lens is placed in a medium of refractive index $4 / 3$ , its focal length will become …… cm.

(1987, 2M)

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Answer:

Correct Answer: 30. 60

Solution:

$\begin{aligned} \frac{1}{f_{air}} & = (1.5 - 1) \frac{1}{R_{1}} - \frac{1}{R_{2}} \\ \frac{1}{f_{medium}} & = \frac{1.5}{4 / 3} - 1 \frac{1}{R_{1}} - \frac{1}{R_{2}} \end{aligned}$

Dividing Eq. (i) by Eq. (ii), we get

$\begin{aligned} \frac{f_{medium}}{f_{air}} & = 4 \\ ∴ f_{medium} & = 4 f_{air} = 4 \times 15 \\ = 60 c m \end{aligned}$

Optics 3 Question 30

30. A thin lens of refractive index 1.5 has a focal length of $15 c m$ in air. When the lens is placed in a medium of refractive index $4 / 3$ , its focal length will become …… cm.

Answer:

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Optics 3 Question 30

30. A thin lens of refractive index 1.5 has a focal length of 15cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become …… cm.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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30. A thin lens of refractive index 1.5 has a focal length of $15 c m$ in air. When the lens is placed in a medium of refractive index $4 / 3$ , its focal length will become …… cm.