Q.1. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio (2010)
Ans. 6
Solution.
Q.2. A large glass slab (μ= 5 / 3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R? (2010)
Ans. 6
Solution. In the figure, C represents the critical angle
Q.3. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 25/3 m to 50/7 m in 30 seconds. What is the speed of the object in km per hour? (2010)
Ans. 3
Solution. Using mirror formula for first position
Using mirror formula for the second position
Change in position of object = 25 m
Q.4. Water (with refractive index = 4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is (2011)
Ans. 2
Solution. For the convex spherical refracting surface of oil we apply
∴ v = 21 cm
For wateroil interface
∴ V' = 16 cm.
This is the image distance from wateroil interface.
Therefore the distance of the image from the bottom of the tank is 2 cm.
Q.5. A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maximum on the surface of water are given by x^{2} = p^{2}m^{2}λ^{2} – d^{2}, where l is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is (JEE Adv. 2015)
Ans. 3
Solution. For maxima
Path defference = mλ
∴ S_{2}A – S_{1}A = mλ
Q.6. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M_{1}. When the setup is kept in a medium of refractive index 7/6, the magnification becomes M2. The magnitude is (JEE Adv. 2015)
Ans. 7
Solution. Applying mirror formula
For convex lens u = 2f_{ℓ}
Therefore image will have a magnification of 1.
When the set – up is kept in a medium
The focal length of the lens will change
⇒ = 17.5 cm.
Applying lens formula
= Magnification by lens
Q.7. The monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n = √3 the value of q is 60° and
The value of m is (JEE Adv. 2015)
Ans. 2
Solution. Here ∠MPQ + ∠MQP = 60°. If ∠MPQ = r then ∠MQP
= 60 – r
Applying Snell’s law at P
sin60° = n sin r ...(i)
Differentiating w.r.t ‘n’ we get
...(ii)
Applying Snell’s law at Q
sin θ = n sin (60° – r) ...(iii)
Differentiating the above equation w.r.t ‘n’ we get
[from (ii)]
[sin (60° – r) + cos (60° – r) tan r] ...(iv)
From eq. (i), substituting n = √3 we get r = 30°
From eq (iii), substituting n = √3 , r = 30° we get θ = 60°
On substituting the values of r and θ in eq (iv) we get
[sin 30° + cos 30° tan 30°] = 2
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
88 docs49 tests
