“You can chain me, you can torture me, you can even destroy this body, but you will never imprison my mind.”
– Mahatma Gandhi
Today we will discuss two questions from optics which appeared in IIT-JEE 2011 question paper. The first one is single correct answer type multiple choice question where as the second one is integer answer type in which the answer is a single digit integer ranging from 0 to 9.
(1) A light ray traveling in glass medium is incident on glass air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is
You can easily rule out sketches (A) and (B) since they indicate 100% transmitted intensity at angle of incidence (θ) of 0º.
[Even at normal incidence (θ = 0) the entire incident light is not transmitted. A small portion is reflected back].
Sketch (C) is the correct option since it indicates total reflection (total internal reflection) at an angle of incidence equal to the critical angle for the glass air interface. The entire light is totally reflected abruptly and the transmitted intensity drops abruptly to zero.
[The gradual change in intensity indicated in sketch (D) is ruled out]. 
(2) 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
The law of distances in the case of refraction at an interface between two media of refractive indices n1 and n2 is
n2/v – n1/u = (n2 – n1)/R where ‘v’ is the image distance, ‘u’ is the object distance and ‘R’ is the radius of curvature of the refracting surface. In the case of the refraction at the air-oil interface we have n2 = 7/4, n1 = 1, u = – 24 cm and R = 6 cm. so that
7/4v – 1/(– 24) = [(7/4) –1]/6
[Note that we have applied the Cartesian sign convention which you can find here].
Therefore, 7/4v = 3/24 – 1/24 = 1/12
This gives v = 21 cm.
The image formed (at 21 cm from the oil surface) by the refraction at the air-oil interface will act as a virtual object for the refraction at the oil-water interface so that u = 21 cm (positive according to Cartesian sign convention). In this case n2 = 4/3, n1 = 7/4 and R = ∞.
[The radius of curvature is infinity since the oil-water interface is plane].
Therefore we have
4/3v1 – 7/(4×21) = [(4/3) – (7/4)]/∞ where v1 is the distance of the final image (from the water surface).
Or, 4/3v1 – 7/(4×21) = 0
This gives v1 = 16 cm.
The water column in the tank has a height of 18 cm. Therefore the final image is 2 cm above the bottom of the tank.
Therefore, x = 2
