Consider the following MCQ which is meant for gauging your grasp of the principle of superposition of waves:

**Two laser beams of the same wave length and intensities 9I and I are superposed. The minimum and maximum intensities of the resultant beam are:**

(a) 8I and 10I (b) 0 and 10I (c) 4I and 16I (d) 0 and 16I (e) 4I and 10I(a) 8I and 10I (b) 0 and 10I (c) 4I and 16I (d) 0 and 16I (e) 4I and 10I

Since the intensity is directly proportional to the square of the amplitude, the minimum and maximum intensities are[√(9I) – √I]

Diffraction at single slit deserves your special attention since questions involving it are often seen in Medical and Engineering entrance test papers. Here is a question which appeared in Kerala Engineering Entrance 2006 question paper:

^{2}and [√(9I) + √I]^{2}respectively, which are [2√I]^{2}and [4√I]^{2}. So, the correct option is 4I and 16I.Diffraction at single slit deserves your special attention since questions involving it are often seen in Medical and Engineering entrance test papers. Here is a question which appeared in Kerala Engineering Entrance 2006 question paper:

**The width of a single slit if the first minimum is observed at an angle 2º with a light of wave length 6980 Ǻ is**

(a) 0.2 mm (b) 2×10

(a) 0.2 mm (b) 2×10

^{–5}mm (c) 2×10^{5}mm (d) 2 mm (e) 0.02 mmThe angular position of the first minimum is λ/a. Therefore, λ/a = 2×(π/180), on converting the angle in degrees to the angle in radian. This gives a = 180 λ/2π

= 180×6980×10

^{–10}/2π = 0.02×10

^{–3}m = 0.02 mm.

Here is a question which may easily mislead you:

**In a double slit experiment, the separation between the slits is 1 mm and the distance between the double slit and the screen is 1m. If the slits are illuminated by monochromatic light of wave length 6000 Ǻ, what is the separation between the 2nd dark bands on either side of the central band?**

(a) 0.9 mm (b) 1.2 mm (c) 1.8 mm (d) 3 mm (e) 4.2 mmAs you know, the central band in Young’s double slit experiment is bright. The distance from the centre of this bright band to the centre of the first dark band (on one side) is half a band (fringe) width. The distance from the centre of the first dark band to the centre of the second dark band is one band width. So, the distance between the centre of the central band and the second dark band on one side is one and a half times the band width. The distance between the second dark bands on either side is therefore three times the band width.

(a) 0.9 mm (b) 1.2 mm (c) 1.8 mm (d) 3 mm (e) 4.2 mm

Now, band width, β = λD/d where λ is the wave length of light used, D is the distance of the screen from the double slit and ‘d’ is the separation between the two slits. The answer to the question is 3β = 3 λD/d =3×6000×10

^{–10}×1/(1×10

^{–3}) = 1.8×10

^{–3}m = 1.8 mm.

The following MCQ appeared in

**AIEEE 2004**question paper:**The maximum number of possible interference maxima for slit separation equal to twice the wave length in Young’s double slit experiment is****(a) infinite (b) five (c) three (d) zero**

If ‘d’ is the slit separation (AB in fig.), the path difference between the waves reaching a point P on the screen is BP – AP = dsinθ and the condition to produce the n

d sinθ = nλ = nd/2.

Therefore, n = 2 sinθ. Since ‘n’ must be an integer, the allowed values are 2,1,0, –1 and –2 because the values of sinθ lie between +1 and –1. So, there can be, at the most, 5 bands only (including the central band.

Now, consider the following MCQ involving interference of light reflected from a thin film:

Since the minimum thickness is required, n =1 so that 2μtcosr = λ from which t = λ/2μcosr = 6000×10

(Note that the condition for darkness in the reflected system is 2μtcosr = nλ [and not (2n+1)λ/2] because of the phase change of π (equivalent to a path change λ/2) when reflection occurs at a denser medium for light traveling through a rarer medium).

^{th}maximum (bright fringe) at P isd sinθ = nλ = nd/2.

Therefore, n = 2 sinθ. Since ‘n’ must be an integer, the allowed values are 2,1,0, –1 and –2 because the values of sinθ lie between +1 and –1. So, there can be, at the most, 5 bands only (including the central band.

Now, consider the following MCQ involving interference of light reflected from a thin film:

**A transparent film of refractive index 1.5 is viewed in reflected monochromatic light of wave length 6000 Ǻ. If the angle of refraction in to the film is 60º, what is the smallest thickness of the film to make it appear dark?**

(a) 4×10In the reflected system, the condition for(a) 4×10

^{–4}mm (b) 5×10^{–4}mm (c) 6×10^{–4}mm (d) 7×10^{–4}mm (e) 8×10^{–4}mm*darkness*is**2μtcosr = nλ**where μ is the refractive index, ‘t’ is the thickness (of the film), ‘r’ is the angle of refraction in to the film, ‘n’ is the order of the interference band and λ is the wave length of the light used.Since the minimum thickness is required, n =1 so that 2μtcosr = λ from which t = λ/2μcosr = 6000×10

^{–10}/(2×1.5×0.5) = 4×10^{–7}m = 4×10^{–4}mm.(Note that the condition for darkness in the reflected system is 2μtcosr = nλ [and not (2n+1)λ/2] because of the phase change of π (equivalent to a path change λ/2) when reflection occurs at a denser medium for light traveling through a rarer medium).

The following MCQ which appeared in IIT 2001 entrance test paper is worth noting:

Similarly, resultant intensity at B = I + 4I + 2√I×√(4I)cosπ = I.

The difference between the rresultant intensities is 5I–I = 4I.

Now consider the following question which involves the measurement of refractive index using interference fringe shift:

Therefore we have (μ–1)t = nλ so that μ = (nλ/t) +1 = (4×5893×10

**Two beams of light having intnsities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is**

(a) 2I (b) 4I (c) 5I (d) 7ISince the wave amplitudes are added vectorially to produce the resultant intensity at any point, the resultant intensity at A = I + 4I + 2√I×√(4I)cos(π/2) = 5I. [Note that the amplitude is directly proportional to the square root of intensity].(a) 2I (b) 4I (c) 5I (d) 7I

Similarly, resultant intensity at B = I + 4I + 2√I×√(4I)cosπ = I.

The difference between the rresultant intensities is 5I–I = 4I.

Now consider the following question which involves the measurement of refractive index using interference fringe shift:

**When a thin sheet of transparent material of thickness 4×10**

(a) 1.378 (b) 1.432 (c) 1.523 (d) 1.546 (e) 1.589Before introducing the transparent sheet, the path difference between the beams arriving at the centre of the fringe system is zero. When the sheet of thickness ‘t’ is introduced with its plane normal to one of the beams, a path difference equal to μt – t = (μ–1)t is produced. [Note that a path length ‘t’ in a medium of refractive index μ is equivalent to a path length μt in air because of the shrinkage of the waves].^{–3}mm is placed in the path of one of the interfering beams in Young’s double slit experiment, it is found that the central bright fringe shifts through a distance equal to four fringes. What is the refractive index of the transparent material? (Wave length of light used is 5893 Ǻ).(a) 1.378 (b) 1.432 (c) 1.523 (d) 1.546 (e) 1.589

Therefore we have (μ–1)t = nλ so that μ = (nλ/t) +1 = (4×5893×10

^{–10})/(4×10^{–6}) + 1 = 1.5893.
Excellent help for students of the faculties of Science

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