Two coherent sources have intensity in the ratio of $\frac{100}{1}$. Ratio of (intensity)_max/(intensity)_min is: 
1. $\frac{1}{100}$
2. $\frac{1}{10}$
3.$\frac{10}{1}$
4. $\frac{3}{2}$

If two waves represented by $y_1=4\sin \omega t$ and $y_2=3\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 3}}\right)$ interfere at a point, the amplitude of the resulting wave will be about 

(1) 7

(2) 6

(3) 5

(4) 3.5

Two coherent sources of intensities, I₁ and I₂ produce an interference pattern. The maximum intensity in the interference pattern will be 

(1) I₁ + I₂

(2) $I_1^2+I_2^2$

(3) (I₁ + I₂)²

(4) ${(\sqrt{I_1}+\sqrt{I_2})}^2$

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi }{2}$ at point A and π at point B. Then the difference between the resultant intensities at A and B is 

(1) 2I

(2) 4I

(3) 5I

(4) 7I

If an interference pattern has maximum and minimum intensities in a \(36:1\) ratio, then what will be the ratio of their amplitudes?
1. \(5:7\)
2. \(7:4\)
3. \(4:7\)
4. \(7:5\)

In Young's double slit experiment, if the slit widths are in the ratio \(1:9,\) then the ratio of the intensity at minima to that at maxima will be:
1. \(1\)
2. \(1/9\)
3. \(1/4\)
4. \(1/3\)

In a certain double slit experimental arrangement interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength 6000 Å, the fringe width will be 

(1) 0.5 mm

(2) 1.0 mm

(3) 1.2 mm

(4) 1.5 mm

Two coherent light sources S₁ and S₂ (λ= 6000 Å) are 1mm apart from each other. The screen is placed at a distance of 25 cm from the sources. The width of the fringes on the screen should be 

(1) 0.015 cm

(2) 0.025 cm

(3) 0.010 cm

(4) 0.030 cm

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are nλ and (n + 2) λ respectively, where n is a whole number and λ is the wavelength. Taking the central fringe as zero, what is formed at X

(1) First bright

(2) First dark

(3) Second bright

(4) Second dark

The Young's experiment is performed with the lights of blue (λ = 4360 Å) and green colour (λ = 5460 Å), If the distance of the 4th fringe from the centre is x, then 

(1) x (Blue) = x (Green)

(2) x (Blue) > x (Green)

(3) x (Blue) < x (Green)

(4) $\frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{5460}{4360}$