The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits.

Monjur Alahi

The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is 4 . If one of the slits is closed, the intensity at this point will be

(a) Io

(b) lo/4

(c) Io/2

(d) 4Io.


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

The correct answer is (b) IO/4

EXPLANATION: 

As in the previous explanation the maximum intensity 

I₀ = (√I₁ + √I₂)². Here given that I₁ = I₂ = I (say). 

→I₀ = 4 I  ------- (i) 

If one slit is closed I₀' = I 

→I₀' = I₀/4      {from (i)} 

Hence option (b)

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
(a) In a single slit diffraction experiment, the width of the slit is made double the original width.

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half...

1 Answers 1 views
A Young’s double slit experiment uses a monochromatic source.

(a)  Straight line fringes are formed on screen.

1 Answers 1 views
A thin transparent sheet is placed in front of a Young's double slit. The fringe-width will (a) increase (b) decrease

(c) remain same EXPLANATION:    The fringe width is given as w = Dλ/d where D = distance of the screen from the slits d = distance of slits.  The wavelength λ remains the same...

1 Answers 1 views
If Young's double slit experiment is performed in water, (a) the fringe width will decrease

(a) the fringe width will decrease EXPLANATION:    As the wavelength of light in water will decrease, hence the fringe width will decrease because the fringe width is proportional to the wavelength.

1 Answers 1 views
If the source of light used in a Young's double slit experiment is changed from red to violet,

(b) consecutive fringes will come closer EXPLANATION:    The options (a), (c) and (d) are related to the intensity changes. Since the intensity of light incident on the slits is not changed...

1 Answers 1 views
A Young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.

The correct answer is (3) Straight line Note: If instead of Young's double slit experiment, young’s double hole experiment was given shape would have been hyperbola.

1 Answers 1 views
How will the angular separation and visibility of fringes in Young’s double slit experiment ,

 (i) Angular separation βθ = β/D = λ/d It is independent of D; therefore, angular separation remains unchanged if screen is moved away from the slits. But the actual separation between fringes β = λD/d  increases, so visibility...

1 Answers 1 views
How would the angular separation of interference fringes in Young’s double slit experiment change

Angular separation between fringes  βθ = λ/d where λ = wavelength, d = separation between coherent sources, βθ is independent of distance between the slits and screen; so angular separation (βθ) will remain unchanged.

1 Answers 1 views
In a single-slit diffraction experiment, the width of the slit is made double the original width.

In single slit diffraction experiment fringe width is β = 2λD/d If d is doubled, the width of central maxima is halved. Thus size of central maxima is reduced to half. Intensity...

1 Answers 1 views
How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled?

The fringe width is β = λD/d If D (distance between slits and screen) is doubled, then fringe width will be doubled.

1 Answers 1 views
X