JEE Exam > JEE Notes > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Refraction of Light- 1
DC Pandey Solutions: Refraction of Light- 1 | DC Pandey Solutions for JEE Physics PDF Download
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
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: 
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: 
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: 
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:
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:
Using the equation,
∴ v = 4.0cm
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 
we get, 
v = + 1.4 cm
Now, 
= - 0.0777
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FAQs on DC Pandey Solutions: Refraction of Light- 1 - DC Pandey Solutions for JEE Physics
| 1. What is refraction of light? |  |
Ans. Refraction of light is the phenomenon of bending of light as it passes from one medium to another. It occurs due to the change in speed of light when it enters a different medium, resulting in a change in its direction.
| 2. How does refraction of light occur? | |
Ans. Refraction of light occurs when light passes from one medium to another with a different optical density. The change in optical density causes the light waves to change their speed and direction, resulting in the bending of light.
| 3. What are the factors that affect the amount of refraction? | |
Ans. The amount of refraction of light depends on two factors: the angle of incidence and the refractive index of the media. The angle of incidence is the angle at which the light ray strikes the interface between the two media, while the refractive index is a measure of how much the speed of light changes in different media.
| 4. How is the refractive index calculated? | |
Ans. The refractive index of a medium is calculated by dividing the speed of light in a vacuum by the speed of light in that medium. Mathematically, refractive index (n) = speed of light in vacuum / speed of light in the medium.
| 5. What are some real-life applications of refraction of light? | |
Ans. Refraction of light has several practical applications in our daily lives. Some examples include the working of lenses in glasses and cameras, the formation of rainbows, the bending of light in optical fibers used in communication, and the functioning of prisms in spectacles and spectroscopes.
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