A prism has refracting angle A and its refractive index is cot(A/2). Find the angle of minimum deviation.

180 - 2A
90 - A
180 + 2A
180 - 3A

Correct answer: 180 - 2A

Solution

Using Snell's law at minimum deviation: n = sin((A+D_min)/2)/sin(A/2). With n = cot(A/2) = cos(A/2)/sin(A/2), we get sin((A+D_min)/2) = cos(A/2) = sin(90 - A/2). So (A+D_min)/2 = 90 - A/2, giving A + D_min = 180 - A, so D_min = 180 - 2A.

Related JEE Advanced Physics questions

The image of an object, formed by a plano-convex lens at a distance of 8m behind the lens, is real and is one third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is -
When two waveguides with identical cross-sectional areas but differing numerical apertures, NA₁ and NA₂ (where NA₂ is smaller than NA₁), are connected end-to-end, the resulting numerical aperture of the system is:
A point light source is submerged inside water (refractive index n = 4/3). When the source is at depth H below the surface, the fraction of total light energy that escapes out of the water is 'a'. When the same source is placed at depth 2H, the fraction that escapes is 'b'. Which of the following correctly relates a and b?
A cylindrical bucket of depth 60 cm is partly filled: the bottom third (20 cm) contains a liquid of refractive index 1.5 and the middle third (20 cm) contains oil of refractive index 2. The top third (20 cm) is air. To an observer looking vertically downward from the top, the volumes of air, liquid and oil appear equal. The apparent depth of the bucket as seen by the observer is alpha cm. Find alpha/5.
A point light source S sits at the bottom of a stack of plane-parallel layers. The bottommost layer has refractive index mu0. The n-th layer above the bottom has refractive index mu(n) = mu0 + mu0/(4n - 18). A ray leaves S at an angle of incidence i = 30° measured from the vertical. At the upper surface of which layer does total internal reflection (TIR) first occur?
A converging lens forms a real image of an object on a fixed screen. The lens is then moved at constant speed v0 along its principal axis away from the screen while the object is simultaneously moved to keep the image on the screen at all times. If the image size is twice the object size at some instant, what is the required speed of the object at that instant?

⚔️ Practice JEE Advanced Physics free + battle 1v1 →