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Young's Double Slit Experiment - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Young's double slit experiment -1, Young's double slit experiment- 2 are considered the most difficult concepts.

  • 103 Questions around this concept.

Solve by difficulty

a YDSE: D = 1 m, d = 1 mm and l = 5000 n m. The distance of 100th maxima from the central maxima is:

Concepts Covered - 2

Young's double slit experiment -1

Young's double slit experiment -1-

This experiment is performed by British physicist Thomas Young. He used an arrangement as shown below. In this he used a monochromatic source of light S . He made two pinholes S1 and S2 (very close to each other) on an opaque screen as shown in figure Each source can be considered as a source of coherent light source. 

 

                                                                                     

                                                         

So, we can see that the monochromatic light source ‘s’ kept at a considerable distance from two slits s1 and s2. The arrangement is such that the S is equidistant from S1 and S2. S1 and S2 behave as two coherent sources, as both bring derived from S. 

Let d be the distance between two coherent sources A and B having wavelength λ. A screen XY is placed parallel to opaque screen at a distance D. O is a point on the screen equidistant from A and B. P is a point at a distance x from O

                                                     

                          


From the above figure, we can see that the waves from A and B meet at P. It may be in phase or out of phase depending upon the path difference between two waves, which we will calculate.

                                             \begin{array}{l}{\text { Draw AM perpendicular to BP }} \\ \\ {\text { The path difference } \delta=\mathrm{BP}-\mathrm{AP}} \\ \\ {\text {As we can see that, AP = MP }} \\ \\ {\delta=\mathrm{BP}-\mathrm{AP}=\mathrm{BP}-\mathrm{MP}=\mathrm{BM}} \\ \\ {\text { In right angled } \mathrm{ABM}, \ \ \mathrm{BM}=\mathrm{d} \sin \theta \text { lf } \theta \text { is small, }} \\\\ {.\ \ \ \ \sin \theta=\theta} \\ \\ {\text { The path difference } \delta=\theta . d} \\ \\ {\text { In right angled triangle } \operatorname{COP}_{,} \ \ \ \tan \theta=\mathrm{OP} / \mathrm{CO}=\mathrm{X} / \mathrm{D}} \\ \\ {\text { For small values of } \theta, \tan \theta=\theta} \\ \\ {\text { Thus, the path difference } \delta=\mathrm{xd} / \mathrm{D}}\end{array}

                                                So, the \ path \ differnece \ is = \frac{xd}{D}

Assumption in this experiment - 

1. D> d: Since D > > d, the two light rays are assumed to be parallel, then the path difference,

2.  d/λ >> 1: Often, d is a fraction of a millimetre and λ is a fraction of a micrometre for visible light.

 

For Bright Fringes - 

By the principle of interference, condition for constructive interference is the path difference = nλ

\begin{array}{l}{\frac{xd}{D}=n \lambda} \\ \\ {\text { Here, } n=0,1,2 \ldots \ldots \text { indicate the order of bright fringes }} \\ {\text { So, } x=(\frac{n \lambda D}{d})} \\ \\ {\text { This equation gives the distance of the } n^{\text {th }} \text { bright fringe from the point O. }}\end{array}

 

For Dark fringes -

By the principle of interference, condition for destructive interference is the path difference =   \frac{(2n-1) \lambda}{2}

Here, n = 1,2,3 … indicate the order of the dark fringes.

                                   So, 

                                                                 x = \frac{(2n-1) \lambda D}{2d}

The above equation give the distance of the nth dark fringe from the point O. 

So, we can say that the alternate dark and bright fringe will be obtained on either side of the central bright fringe. 

 

Band Width (β) -

The distance between any two consecutive bright or dark bands is called bandwidth. 

Take the consecutive dark or bright fringe - 

                                           \\ x_{n+1} - x_n= \frac{(n+1) \lambda D}{d} - \frac{(n) \lambda D}{d} \\ \\ x_{n+1} - x_n = \frac{\lambda D}{d}

                                                              \beta = \frac{\lambda D}{d}

Angular fringe width -

 

                                                            \begin{array}{l} \\ \\ {\theta=\frac{\beta}{D}=\frac{\lambda D / d}{D}=\frac{\lambda}{d}}\end{array}

 

 

Young's double slit experiment- 2

Young's double slit experiment- 2 -

Intensity of Fringes In Young’s Double Slit Experiment-

 

For two coherent sources S1 and S2, the resultant intensity at point P on the screen is given by-

                                              I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \phi

where 

I_{1}= The intensity of wave from S1

I_{2}= The intensity of wave S2

Putting I_1 \ and \ I_2 = I_o   (Becasue d<<<D)

                                         \Rightarrow I=I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos \phi=4 I_{0} \cos ^{2} \frac{\phi}{2}          

So the intensity variation from maximum to minimum depends on the phase difference. Let us discuss one by one -

For maximum intensity

The phase difference between the waves at the point of observation is  \phi=0^{\circ} \text { or } 2 n \pi. Path difference between the waves at the point of observation is \Delta x=n \lambda(i . e . \text { even multiple of } \lambda / 2)

Resultant intensity at the point of observation will be maximum

                                                          \begin{aligned} \ \ i.e \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \\ I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \\ \text { If } \quad I_{1} &=I_{2}=I_{0} \Rightarrow I_{\max }=4 I_{0} \end{aligned}

For Minimum Intensity -

The phase difference between the waves at the point of observation is 

                                                                     \begin{array}{l}{\phi=180^{\circ} \text { or }(2 n-1) \pi ; n=1,2, \ldots} \\ {\text { or }(2 n+1) \pi ; n=0,1,2 \ldots .}\end{array}

Path difference between the waves at the point of observation is \Delta x=(2 n-1) \frac{\lambda}{2}(\text { i.e. odd multiple of } \lambda / 2)

Resultant intensity at the point of observation will be maximum

                                                             \begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \\ {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \\ {\text { If } I_{1}=I_{2}=I_{0} \Rightarrow I_{\min }=0}\end{array}

 

 

                                    

 

Maximum Order of Interference Fringes -

 

As we know that the position of nth order maxima on the screen is - 

                                                                           \frac{n \lambda D}{d} ; n=0, \pm 1, \pm 2, \ldots

Value of 'n'  cannot be taken as infinitely large, because it violates the assumption of the Young's double slit experiment (Mentioned in the last concept) which means that the \theta is small or we can write x<<D. So, 

                                                                               \Rightarrow \frac{x}{D}=\frac{n \lambda}{d}<<1

So the above formula is only applicable for  - 

                                                                    n<<\frac{d}{\lambda}

But when,   n \approx \frac{d}{\lambda},    which means that the n is comparable with    \frac{d}{\lambda} . Then the above formula is not applicable, then we have to go with the basic and we will equate path difference as - 

                                                                  \begin{array}{l}{\Rightarrow d \sin \theta=n \lambda} \\ \\ {\Rightarrow n=\frac{d \sin \theta}{\lambda}} \\ \\So, \\ \ \ n_{\max }=\left[\frac{d}{\lambda}\right] \end{array}

                                   The above represents box function or greatest integer function.

Similarly, the highest order of interference minima 

                                                                       n_{\min }=\left[\frac{d}{\lambda}+\frac{1}{2}\right]

 

 

  

 

 

 

Study it with Videos

Young's double slit experiment -1
Young's double slit experiment- 2

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