Case Study: Read the following paragraph and answer the questions.
A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path. The angle of deviation of a prism is, δ = (μ- 1) A, through which a ray deviates on passing through a thin prism of small refracting angle A.
(i) For which colour, angle of deviation is minimum?
(ii) At what angle of incidence is the deviation through a prism maximum?
(iii) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644).
OR
(iii) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle of prism is 60°, then what is the angle of minimum deviation?

Question

Case Study: Read the following paragraph and answer the questions.
A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path. The angle of deviation of a prism is, δ = (μ- 1) A, through which a ray deviates on passing through a thin prism of small refracting angle A.
(i) For which colour, angle of deviation is minimum?
(ii) At what angle of incidence is the deviation through a prism maximum?
(iii) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644).
OR
(iii) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle of prism is 60°, then what is the angle of minimum deviation?

Infinity Learn · Accepted Answer

If μ is refractive index of the material of the prism, then prism formula is,

$μ = \frac{\sin (A + δ_{m}) / 2}{\sin A / 2}$

(i) Angle of deviation is minimum for the red colour.

(ii) The deviation is maximum when angle is 90°.

(iii) A = 6⁰; μ = 1.644

$f = (μ - 1) A f = (1.644 - 1) 6 = 0.644 \times 6 δ = 3 . 864^{0}$

OR

$A + δ_{m} = i_{1} + i_{2} = 50^{\circ} + 50^{\circ} = 100^{\circ}$

(i) (b) The I - V characteristics of an LED is similar to that of a Si junction diode. But the threshold voltages are much higher and slightly different for each colour.

(ii)

(iii) $As E_{g} = \frac{hc}{λ} ∴ λ = \frac{hc}{E_{g}}$

$Here, E_{g} = 1.9 eV, he = 1240 eV nm ∴ λ = \frac{1240 eVnm}{1.9 eV} = 652.6 nm$

OR

(iii) LEDs are made up of compound semiconductors and not by the elemental conductor because the band gap in the elemental conductor has a value that can detect the light of a wavelength which lies in the infrared (IR) region.

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