Apparent Depth of Object Numerical Class-12 Nootan ISC Physics Solution Ch-15 Refraction of Light at a Plane Interface : Total Internal Reflection : Optical Fiber. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Apparent Depth of Object Numerical Class-12 Nootan ISC Physics Solution Ch-15 Refraction of Light at a Plane Interface : Total Internal Reflection : Optical Fiber
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-15 | Refraction of Light at a Plane Interface |
| Topics | Numericals on Apparent Depth of Object |
| Academic Session | 2025-2026 |
Numericals on Apparent Depth of Object
Class-12 Nootan ISC Physics Solution Ch-15 Refraction of Light at a Plane Interface : Total Internal Reflection : Optical Fiber.
Que-4: How deep will a 4 m deep tank appear when seen in air due to optical refraction? The refractive index of water is 4/3.
Ans- μ = real depth / apparent depth
=> 4/3 = 4 / apparent depth
=> apparent depth = 3 m
Que-5: A beaker is filled with water to a height of 12.5 cm. The apparent depth of a needle at the bottom of the beaker is 9.4 cm. Find refractive index of water.
Ans- μ = real depth / apparent depth
=> μ = 12.5 / 9.4 = 1.33
Que-6: The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth?
Ans- μ = real depth / apparent depth
=> 1.3 = real depth / 7.7
=> real depth = 1.3 x 7.7 = 10 cm
Que-7: A needle lying on horizontal surface is viewed vertically from a distance of 100 cm. A glass slab of 12 cm thickness is placed on the needle. Find the distance of the vertical image of the needle from the eye, taking ang = 1.5.
Ans- t = 12 cm
vertical shift = t (1 – 1/ang)
= 12 (1 – 1/1.5) = 4 cm
∴ distance of image = 100 – 4 = 96 cm
Que-8: A microscope focused on a pin lying at the bottom of a beaker reads 3.965 cm. When a liquid is poured up to a height of 2.537 cm into the beaker, the microscope focused again on the pin reads 3.348 cm. Find the refractive index of the liquid.
Ans- vertical shift = t (1 – 1/μ)
=> 3.965 – 3.348 = 2.537 (1 – 1/μ)
=> 0.617 / 2.537 = 1 – 1/μ
=> 1/μ = 0.7568
=> μ = 1.321
Que-9: A beaker is filled with a liquid to a height of 14 cm. The apparent depth of a needle fixed at the bottom of the beaker is measured to be 10 cm by a microscope. What is the refractive index of the liquid? The height of the liquid in the beaker is now raised to 21 cm. By what distance would the microscope have to be moved to focus on the needle again?
Ans- n = real depth / apparent depth
=> n = 14/10 = 1.4
new apparent depth
= 21/1.4 = 15 cm
∴Vertical shift of microscope = 15 – 10 = 5 cm
Que-10: A beaker 20 cm high is half filled with an oil of refractive index 1.40 and the upper half filled with water of refractive index 1.33. What is the apparent depth of the beaker as observed normally?
Ans- apparent depth by two liquids
= 10/1.40 + 10/1.33 = 14.66 cm
Que-11: A beaker is filled with water (n = 1.5) to a height h. Height of the beaker is 15 cm. When the bottom of the beaker is viewed vertically by an observer, beaker appears to be half filled. Find the value of h.
Ans- n = real depth / apparent depth
=> 1.5 = h/(15 – h)
=> 22.5 – 1.5h = h
=> 2.5 h = 22.5
=> h = 9 cm.
Que-12: A container of uniform cross-section has a height of 14 m. Up to what height water of refractive index 4/3 should be filled inside the container so that container seems to be half filled for normal viewing.
Ans- apparent depth = real depth / μ
= x / (4/3) = 3x/4
Length of the air column will be(14−x) . As the container must appear to be half-filled, the apparent depth will be equal to the length of the air column.
=> 3x/4 = 14 – x
=> x = 8 m
— : End of Apparent Depth of Object Numerical Class-12 Nootan ISC Physics Solution Ch-15 :–
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