Question

Q 52 A convex lens of focal length 0.2m is made of material of refractive index 7/2. The lens is placed in a medium, then its focal length is increased by 0.35 m.
What is the refractive index of the medlum?
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Answer 1

Answer:

equal to the speed of light (c) divided by the velocity of light through the medium (v).

Explanation:

The refractive index of a medium (n) is equal to the speed of light (c) divided by the velocity of light through the medium (v).

hope it's helpful

Answer 2

Explanation:

• So focal length in air is fa = 0.2 m
• Also μ = 7/2
•      Now μ / μa = 7/2 / 1 = 7/2
• So the refractive index of liquid = μL
• Now initially focal length of liquid is
• fL = 0.2 m + 0.35 m
•     = 0.55 m
• So we have the lens makers formula we gt
•           1/fa = (μ / μa – 1) [1/R1 – 1/R2] ------------1
•            1/fL = (μ/μL – 1) [1/R1 – 1/R2] -------------2
• So eqn 1 / 2 we get
•            So fL / fa = (μ/μa – 1) / (μ/μL – 1)
•                     0.55 / 0.2 = (7/2 – 1) / (7/2μL – 1)
•                         11/4 = (7/2 – 1) / (7/2μL – 1)
•            11/4 x (7/2μL – 1) = 5/2
•                 7/2μL – 1 = 5/2
•                7/2μL – 1 = 10/11
•                    7/2μL = 21/11
•                   Or μL = 11/6

Reference link will be

https://brainly.in/question/47806394