CBSE Class 12 Physics Refraction at Plane Surface Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.
Ex.1 The apparent depth of a swimming pool is 1.2 m . What is its real depth ?
Sol. Swimming pool means μwater = 4/3. So that the real depth
R = a μ
= 1.2 x 4/ 3
= 1.6 m
Ex.2 A glass plate 4 mm thick is viewed from the above through a microscope. The microscope must be lowered 2.58 mm as the operator shifts from viewing the top surface to viewing the bottom surface through the glass. What is the index of refraction of the glass ?
Sol. From the information given, it is clear that the apparent depth is 2.58 mm and the real depth is 4mm. Therefore, the refractive index will be
μ = R/ a = 4/ 2.58 = 1.55
Ex.3 A vertical microscope is focussed on a point at the bottom of an empty tank. Water (μ = 4/3) is then poured into the tank. The height of the water column is 4cm. Another lighter liquid, which does not mix with water and which has refractive index 3/2 is then poured over the water.The height of liquid column is 2cm. What is the vertical distance through which the microscope must be moved to bring the object in focus again ?
Sol. The apparent shift of the bottom point upwards will be
Ex.4 Light from a sodium (λ0 = 589 nm) passes through a tank of glycerin (refractive index 1.47) 20m long in a time t1. If it takes a time t2 to traverse the same tank when filled with carbon disulphide (index 1.63), then the difference t2 − t1 is
(A) 6.67 x 10−8 sec (B) 1.09 x 10−7 sec (C) 2.07 x 10−7 sec (D) 1.07 x 10−7 sec
Ex.5 A ray of light falls on a transparent glass slab of refractive index μ. If the reflected ray and refracted ray are mutually perpendicular, then the angle of incidence is
(A) sin−1 (μ) (B) sin−1 (1/μ) (C) tan−1 (1/μ) (D) tan−1 (μ)
Sol. When the reflected ray and refracted ray are mutually perpendicular , then