in a double slit setup, the width of the slits is 0.01 mm and the separation between the slits is 0.05 mm. the number of complete fringes formed between the first minima of the envelope on either side of the central maxima is

get in a double slit setup, the width of the slits is 0.01 mm and the separation between the slits is 0.05 mm. the number of complete fringes formed between the first minima of the envelope on either side of the central maxima is from screen.

In a Young's double slit experiment, the separation between the slits is 0.15 mm . In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

Click here👆to get an answer to your question ✍️ In a Young's double slit experiment, the separation between the slits is 0.15 mm . In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

Question

In a Young's double slit experiment, the separation between the slits is 0.15mm. In the experiment, a source of light of wavelength 589nm is used and the interference pattern is observed on a screen kept 1.5m away. The separation between the successive bright fringes on the screen is :

A

3.9mm

B

6.9mm

C

5.9mm

D

4.9mm

Hard JEE Mains Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

Fringe width  w=

d Dλ ​ where

D= distance b/w screen and slit =1.5m

λ= wavelength =589×10

−9 m

d= distance between slit

=0.15×10 −3 m=1.5×10 −4 m w= 1.5×10 −4 1.5×589×10 −9 ​ w=5.89mm w≈5.9mm

Distance between successive bright fringes is equal to fringe width =w

=5.9mm

option (C) is corrected.

Solve any question of Wave Optics with:-

Patterns of problems

>

Was this answer helpful?

15 0

स्रोत : www.toppr.com

In Young's double slit experiment, the separation between the slits is 0.1 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1 m away from the slits. Find the separation between the successive bright fringes.

In Young's double slit experiment, the separation between the slits is 0.1 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1 m away from the slits. Find the separation between the successive bright fringes.

Byju's Answer Standard XII Physics

YDSE Vs Single Slit Diffraction

In Young's do... Question

In Young's double slit experiment, the separation between the slits is

0.1 mm

, the wavelength of light used is

600 nm

and the interference pattern is observed on a screen

1 m

away from the slits. Find the separation between the successive bright fringes.

A 2 mm B 6 mm C 4 mm D 3 mm Open in App Solution

The correct option is B

6 mm Given: d = 0.1 mm = 10 − 4 m λ = 600 nm = 6 × 10 − 7 m D = 1 m

Now, the separation between the successive bright fringes is,

β = λ D d = 6 × 10 − 7 × 1 10 − 4 = 6 mm Hence, ( B )

is the correct answer.

Why this question?

Tip : Separation between successive bright fringes

and dark fringes is equal to the fringe width.

Suggest Corrections 4

SIMILAR QUESTIONS

Q. In a Young's double slit experiment, the separation between the slits is

0.15 mm

. In the experiment, a source of light of wavelength

589 nm

is used, and the interference pattern is observed on a screen kept

1.5 m

away. The separation between the successive bright fringes on the screen is:

Q. In young's double slit experiment, interference pattern is observed on the screen

L

distance away from slits, distance between centre of two consecutive bright fringes is

x

and slits separation is

d

, then the wavelength of light will be –

Q. A double slit experiment is performed with sodium light of wavelength

600 nm

and interference pattern is observed on a screen

100 cm

away from the slits. The

4 th

dark fringe is at a distance of

7 mm

from the central maximum. Find the separation between the slits.

Q. In Young's double slit experiment using monochromatic light of wavelength

600

nm, interference pattern was obtained on a screen kept

1.5

m away from the plane of the two slits. Calculate the distance between the two slits, if fringe separation/fringe width was found to be

1.0 mm.

Q. In a double slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be seen on the screen one due to light with wavelength 480 nm, and the other due to light with wavelength 600 nm. What is the separation on the screen between the fifth order bright fringes of the two interference patterns?

View More RELATED VIDEOS

Diffraction II PHYSICS Watch in App EXPLORE MORE

YDSE Vs Single Slit Diffraction

Standard XII Physics

स्रोत : byjus.com

4.3 Double

This free textbook is an OpenStax resource written to increase student access to high-quality, peer-reviewed learning materials.

University Physics Volume 3

4.3 Double-Slit Diffraction

4.3 Double-Slit Diffraction

Learning Objectives

By the end of this section, you will be able to:

Describe the combined effect of interference and diffraction with two slits, each with finite width

Determine the relative intensities of interference fringes within a diffraction pattern

Identify missing orders, if any

When we studied interference in Young’s double-slit experiment, we ignored the diffraction effect in each slit. We assumed that the slits were so narrow that on the screen you saw only the interference of light from just two point sources. If the slit is smaller than the wavelength, then Figure 4.10(a) shows that there is just a spreading of light and no peaks or troughs on the screen. Therefore, it was reasonable to leave out the diffraction effect in that chapter. However, if you make the slit wider, Figure 4.10(b) and (c) show that you cannot ignore diffraction. In this section, we study the complications to the double-slit experiment that arise when you also need to take into account the diffraction effect of each slit.

To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets, and then we sum the wavelets from all the slits. This gives the intensity at any point on the screen. Although the details of that calculation can be complicated, the final result is quite simple:

TWO-SLIT DIFFRACTION PATTERN

The diffraction pattern of two slits of width a that are separated by a distance d is the interference pattern of two point sources separated by d multiplied by the diffraction pattern of a slit of width a.

In other words, the locations of the interference fringes are given by the equation

dsinθ=mλ dsinθ=mλ

, the same as when we considered the slits to be point sources, but the intensities of the fringes are now reduced by diffraction effects, according to Equation 4.4. [Note that in the chapter on interference, we wrote

dsinθ=mλ dsinθ=mλ

and used the integer m to refer to interference fringes. Equation 4.1 also uses m, but this time to refer to diffraction minima. If both equations are used simultaneously, it is good practice to use a different variable (such as n) for one of these integers in order to keep them distinct.]

Interference and diffraction effects operate simultaneously and generally produce minima at different angles. This gives rise to a complicated pattern on the screen, in which some of the maxima of interference from the two slits are missing if the maximum of the interference is in the same direction as the minimum of the diffraction. We refer to such a missing peak as a missing order. One example of a diffraction pattern on the screen is shown in Figure 4.11. The solid line with multiple peaks of various heights is the intensity observed on the screen. It is a product of the interference pattern of waves from separate slits and the diffraction of waves from within one slit.

Figure 4.11 Diffraction from a double slit. The purple line with peaks of the same height are from the interference of the waves from two slits; the blue line with one big hump in the middle is the diffraction of waves from within one slit; and the thick red line is the product of the two, which is the pattern observed on the screen. The plot shows the expected result for a slit width

a=2λ a=2λ and slit separation d=6λ d=6λ . The maximum of m=±3 m=±3

order for the interference is missing because the minimum of the diffraction occurs in the same direction.

EXAMPLE 4.3

Intensity of the Fringes

Figure 4.11 shows that the intensity of the fringe for

m=3 m=3

is zero, but what about the other fringes? Calculate the intensity for the fringe at

m=1 m=1 relative to I 0 , I0,

the intensity of the central peak.

Strategy

Determine the angle for the double-slit interference fringe, using the equation from Interference, then determine the relative intensity in that direction due to diffraction by using Equation 4.4.

Solution

From the chapter on interference, we know that the bright interference fringes occur at

dsinθ=mλ dsinθ=mλ , or sinθ= mλ d . sinθ=mλd. From Equation 4.4, I= I 0 ( sinβ β ) 2 ,whereβ= ϕ 2 = πasinθ λ .

I=I0(sinββ)2,whereβ=ϕ2=πasinθλ.

Substituting from above,

β= πasinθ λ = πa λ ⋅ mλ d = mπa d .

β=πasinθλ=πaλ·mλd=mπad.

For a=2λ a=2λ , d=6λ d=6λ , and m=1 m=1 , β= (1)π(2λ) (6λ) = π 3 . β=(1)π(2λ)(6λ)=π3.

Then, the intensity is

I= I 0 ( sinβ β ) 2 = I 0 ( sin(π/3) π/3 ) 2

स्रोत : openstax.org