JEE > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Refraction of Light- 1

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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

Sol:

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.

Sol:

Solving we get v = -9.0 cm

Sol:

Using the equation,

∴ v = 4.0cm

Sol:

we get,

v = + 1.4 cm

Now,

= - 0.0777

The document DC Pandey Solutions: Refraction of Light- 1 | DC Pandey Solutions for JEE Physics is a part of the JEE Course DC Pandey Solutions for JEE Physics.

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