In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other…

Q: In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern

(A) the intensities of both the maxima and the minima increases

(B) the intensity of maxima increases and the minima has zero intensity

(C) the intensity of maxima decreases and that of minima increases

(D) the intensity of maxima decreases and the minima has zero intensity

Ans: (A)

Sol: In normal conditions , when widths of both the slits are equal.

I1 = I2 = I

$\large I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = 4 I$

$\large I_{min} = (\sqrt{I_1} – \sqrt{I_2})^2 = 0$

When width of one of the slit is increased , Intensity due to that slit will be increases while that of other will remain same .

I1 = n I & I2 = I (here n> 1)

$\large I_{max} = I(\sqrt{n}+1)^2 > 4 I $

$\large I_{min} = I(\sqrt{n}-1)^2 > 0 $

intensities of both the maxima and the minima increases .

Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm…

Q: Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, then the observed pattern will reveal

(A) that the central maximum is narrower

(B) more number of fringes

(C) less number of fringes

(D) no diffraction pattern

Ans: (D)

Sol: Diffraction is obtained when slit width is of order of wavelength of light used. The wavelength of X-rays (1 – 100 A°) << slit width (0.6 mm) . Hence no diffraction is observed .

A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is…

Q: A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is

(A) zero

(B) π/2

(C) π

(D) 2π

Ans: (D)

Sol: $\large path \; diff. \Delta x = a sin\theta $

$\large \Delta x = a \frac{y}{D} $

For first minima $\large asin\theta = \lambda $

$\large a \frac{y}{D} = \lambda $

$\large \Delta x = \lambda $

Phase difference $\large = \frac{2\pi}{\lambda} \times \lambda$

= 2π

Two coherent monochromatic light beams of intensities I and 4I are superposed…

Q: Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

(A) 5I and I

(B) 5I and 3I

(C) 9I and I

(D) 9I and 3I.

Ans: (C)

Sol: using formula

$\large I = I_1 + I_2 + 2 \sqrt{I_1 I_2 } cos\phi $

I1 = I , I2 = 4I

$\large I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ , when φ = 0°

$\large I_{max} = (\sqrt{I} + \sqrt{4I})^2 = 9 I$

$\large I_{min} = (\sqrt{I_1} – \sqrt{I_2})^2$ , when φ = 180°

$\large I_{min} = (\sqrt{I} – \sqrt{4I})^2 = I$