In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves having same intensity is 1/4th of their wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fri

Question

In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves having same intensity is $\frac{1}{4}$ th of their wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is

Smart Achievers · Accepted Answer

(2)

$Δ ϕ = \frac{π}{2}$ $\frac{I}{I_{\max}} = \frac{I_{0} + I_{0}}{4 I_{0}} = \frac{1}{2}$

Solution

Q.
In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves having same intensity is $\frac{1}{4}$ th of their wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is

1 $\frac{1}{4}$

2 $\frac{1}{2}$

3 $\frac{1}{6}$

4 $\frac{1}{8}$

Ans.
(2)

$Δ ϕ = \frac{π}{2}$ $\frac{I}{I_{\max}} = \frac{I_{0} + I_{0}}{4 I_{0}} = \frac{1}{2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. In a double-slit experiment, at a certain point on the screen the path difference between the two interfering waves having same intensity is 14th of their wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is

1 14

2 12

3 16

4 18

Ans. (2) Δϕ=π2 IImax=I0+I04I0=12