Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Introductory Exercise 28.1

Ques 1: There is a dust particle on a glass slab of thickness t and refractive index μ = 1.5. When seen from one side of the slab, the dust particle appears at a distance 6 cm. From other side it appears at 4 cm. Find the thickness t of the glass slab.
Sol: 
Actual distance from one side = μ - times = 6 × 1.5 = 9 cm
From other side = 4 × 1.5 = 6 cm
∴ Total thickness = (9 + 6) cm = 15 cm

Ques 2: Given that 1μ2 = 4/3, 2μ3 = 3/2. Find 1μ3.
Sol: 
1μ2 × 2μ3 × 3μ1 = 1
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 3: A monochromatic light beam of frequency 6.0 × 1014 Hz crosses from air into a transparent material where its wavelength is measured to be 300 nm. What is the index of refraction of the material?
Sol: 
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Introductory Exercise 28.2

Ques 1: A glass sphere (μ = 1.5) with a radius of 15.0 cm has a tiny air bubble 5 cm above its centre. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere?
Sol: 
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Solving we get v = - 8.57 cm

Ques 2: One end of a long glass rod (μ = 1.5) is formed into a convex surface of radius 6.0 cm. An object is positioned in air along the axis of the rod. Find the image positions corresponding to object distances of
(a) 20.0 cm (b) 10.0 cm (c) 3.0 cm from the end of the rod.
Sol: 

Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Solving we get v = + 45 cm Similarly other parts can be solved.

Ques 3: A dust particle is inside a sphere of refractive index  4/3. If the dust particle is 10.0 cm from the wall of the 15,0 cm radius bowl, where does it appear to an observer outside the bowl.
Sol:

Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Solving we get v = -9.0 cm

Ques 4: A parallel beam of light enters a clear plastic bead 2.50 cm in diameter and index 1.44. At what point beyond the bead are these rays brought to a focus?
Sol:

Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Using the equation,
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴  v = 4.0cm
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 5: The left end of a long glass rod of index 1.6350 is grounded and polished to a convex spherical surface of radius 2.50 cm. A small object is located in the air and on the axis 9.0 cm from the vertex. Find the lateral magnification.
Sol:
Using Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
we get, Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
v = + 1.4 cm
Now, Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
Refraction of Light (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev
= - 0.0777



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