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A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is
- a)∞
- b)3 m
- c)6 m
- d)4 m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A disc is placed on a surface of pond which has refractive index 5/3. ...
We need to block the light rays using the disc till total internal reflection takes place.
As shown in figure, the direction of light ray at the edge of disc is parallel to light surface, when the incident ray makes critical angle
After the edge, light rays will be reflected due to total internal reflection and the whole radiation from light source is blocked
Hence the angle i shown in figure should be critical angle.
For critical angle we have, sin i = 1/μ = 3/5
hence tan i = 3/4 ....(1)
From figure, we see tan i = r/4 ....(2)
hence from (1) and (2), the required radius = 3 m
Most Upvoted Answer
A disc is placed on a surface of pond which has refractive index 5/3. ...
We need to block the light rays using the disc till total internal reflection takes place.
As shown in figure, the direction of light ray at the edge of disc is parallel to light surface, when the incident ray makes critical angle
After the edge, light rays will be reflected due to total internal reflection and the whole radiation from light source is blocked
Hence the angle i shown in figure should be critical angle.
For critical angle we have, sin i = 1/μ = 3/5
hence tan i = 3/4 ....(1)
From figure, we see tan i = r/4 ....(2)
hence from (1) and (2), the required radius = 3 m
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Community Answer
A disc is placed on a surface of pond which has refractive index 5/3. ...
Given Data:
- Refractive index of liquid (n) = 5/3
- Distance of light source below the surface of liquid (h) = 4 m
Calculation:
- Let the minimum radius of the disc be 'R'.
- The critical angle of incidence for total internal reflection can be calculated using the formula:
sin(critical angle) = 1/n
- The critical angle is the angle of incidence at which light undergoes total internal reflection.
- In this case, the critical angle is given by sin(critical angle) = 1/(5/3) = 3/5
- Let the angle of incidence at the edge of the disc be 'θ'.
- Using trigonometry, sin(θ) = R/(R^2 + h^2), where R^2 + h^2 is the hypotenuse of the right triangle formed by the radius of the disc and the depth of the light source.
- For total internal reflection to occur, θ should be greater than or equal to the critical angle.
- Therefore, we have sin(θ) >= 3/5.
- Substituting the value of sin(θ) from the above equation, we get R/(R^2 + h^2) >= 3/5.
- Simplifying the inequality, we get R^2 + h^2 >= 5R/3.
- Substituting the value of 'h' and solving the inequality, we get R^2 - (5/3)R + 4 >= 0.
- This inequality will be satisfied for R >= 3 m.
Therefore, the minimum radius of the disc needed so that light is not coming out is 3 m.
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A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer?.
A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out isa)∞b)3 mc)6 md)4 mCorrect answer is option 'B'. Can you explain this answer?.
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