In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

\\(d = 10^{-3}\:m\\)\\(D=1\:m\\)\\( \lambda = 500 \times 10^{-9}\:m\\)The width of central maxima in a single slit diffraction: \\(\dfrac{2 \lambda D}{a}\\)Fringe width in double slit pattern: \\(\beta = \dfrac{\lambda D}{d}\\)Given,\\(10 \beta = \dfrac{2 \lambda D}{a}\\)\\( \Rightarrow 10 \dfrac{\lambda D}{d} = \dfrac{2 \lambda D}{a}\\)\\( \Rightarrow a =\dfrac{d}{5}\:mm= 0.2\:mm\\)

A beam of light consisting of two wavelengths, \\(650 nm\\) and \\(520 nm\\), is used to obtain interference fringes in a Young's double-slit experiment.(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength \\(650 nm\\).(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the two slits is \\(0.28mm\\) and the screen is at a distance of \\(1.4m\\) from the slits.

GivenThe wavelength of the light is \\( \lambda_1 = 650 nm\\)The wavelength of seond light,\\( \lambda_2 = 520 nm\\)Distance between the slit and the screen is \\(1.4m\\).Distance between the slits is \\(0.28mm\\).\\((a)\\)The relation between the \\(n^{th}\\) bright fringe and the width of fringe is:\\(x = n\lambda_1\dfrac Dd\\)For third bright fringe, \\(n = 3\\)\\( x = 3 \times 650\dfrac {1.4}{0.28\times10^{-3}}= 1950 \times5\times10^3 nm\\)\\(x=9.75\times10^{-3} m\\)\\(=9.75mm\\)\\((b)\\)We can consider that \\(n^{th}\\) bright frind of \\(\lambda_2\\) and the \\((n-1)^{th}\\) bright fringe of wavelength \\(\lambda_1\\) coincide with each other.\\(n\lambda_2 = (n - 1) \lambda_1\\)\\(520n = 650n - 650\\)\\(650 = 130n\\)\\( n = 5\\)therefore, the least distance from the central maximum can be obtained as:\\(x' = n\lambda_2 \dfrac Dd \\)\\(x'= 5 \times 520 \dfrac Dd = 2600 \dfrac {1.4}{0.28\times10^{-3}} nm\\)\\(x'=1.30\times10^{-2}m=1.3cm\\)

The maximum intensity in Young's double slit experiment is \\({I}_{0}\\). Distance between the slits is \\(d=5\lambda\\), where \\(\lambda\\) is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance \\(D=10d\\)?

Path difference \\(=\Delta x=\dfrac{xd}{D}\\)-------(i)In front of one of the slits,\\(x=\dfrac{d}{2}\\) but \\(d=5\lambda\\)\\(x=\dfrac{5\lambda}{2}\\) and \\(D=10d=50\lambda\\)Now, from equation (i), we have:\\(\Delta x=\dfrac{5\lambda}{2}\times\dfrac{5\lambda}{50\lambda}=\dfrac{\lambda}{4}\\)So, corresponding phas defference:\\(\phi=\dfrac{2\pi}{\lambda}.(\Delta x)=\dfrac{2\pi}{\lambda}\times\dfrac{\lambda}{4}=\dfrac{\pi}{2}\\)As, \\(I=I_0\cos^2(\dfrac{\phi}{2})\\)So, \\(I=I_0\cos^2(\dfrac{\pi}{4})=\dfrac{I_0}{2}\\)

Two coherent monochromatic light beams of intensities \\(I\\) and \\(4I\\) are superposed. The maximum and minimum possible resulting intensities are :

Maximum intensity, \\(I_{max} = (\sqrt {I_{1}} + \sqrt {I_{2}})^{2}\\)\\(= (\sqrt {I} + \sqrt {4I})^{2}\\)\\(= 9I\\)Minimum intensity \\(I_{min} = (\sqrt {I_{1}} - \sqrt {I_{2}})^{2}\\)\\((\sqrt {I} - \sqrt {4I})^{2}\\)\\(= I\\)

In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Distance between the slits, \\(d=0.28\times10^{-3}\ m\\)Distance between the slits and the screen, \\(D = 1.4 m\\)Distance between the central fringe and the fourth \\((n = 4)\\) fringe, \\(u=1.2\times10^{-2}\ m\\)In case of a constructive interference, we have the relation for the distance between the two fringes as:\\(u=n\lambda D/d\\)\\(\Rightarrow \lambda=ud/nD=6\times10^{-7}m=600nm\\)

The interference pattern is obtained with two coherent light source of intensity ratio n. In the interference pattern, the ratio \\(\dfrac{I_{max}-I_{min}}{I_{max}+I_{min}}\\) will be

Given : \\(I_2 = n I_1\\)Maximum intensity of interference \\(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\\)\\(\therefore\\) \\(I_{max} = (\sqrt{I_1} + \sqrt{nI_1})^2 = (1+\sqrt{n})^2 I_1 = (1+ n +2\sqrt{n } ) I_1\\)Minimum intensity of interference \\(I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2\\)\\(\therefore\\) \\(I_{min} = (\sqrt{I_1} - \sqrt{nI_1})^2 = (1-\sqrt{n})^2 I_1 = (1+ n - 2\sqrt{n})I_1\\)\\(\therefore\\) \\(\dfrac{I_{max} -I_{min}}{I_{max} + I_{min}} = \dfrac{2\sqrt{n} - (-2\sqrt{n})}{2(1+n) } = \dfrac{2\sqrt{n}}{1+n}\\)

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for the maximum and zero intensity.

Let \\(y_1\\) and \\(y_2\\) be displacement of two waves having same amplitude a and phase difference \\(\phi\\) between them. \\(y_1=a\sin\omega t\\)\\(y_2=a\sin(\omega t+\phi)\\)Resultant displacement is: \\(y=y_1+y_2\\)\\(y=a\sin\omega t+a\sin(\omega t+\phi)=a\sin\omega t(1+\cos\phi)+\cos\omega t(a\sin\phi) \\)\\(R\cos \theta=a(1+\cos\phi)\\)\\(R\sin \theta=a\sin\phi\\)\\(y=R\sin(\omega t +\theta)\\)Where, \\(R\\) is resultant amplitude at \\(P\\), \\(I\\) is intensity, squaring the equations we get,\\(I=R^2=a^2(1+\cos\phi)^2+a^2(\sin\phi)^2=2a^2(1+\cos\phi)=4a^2\cos^2\dfrac{\phi}{2}\\)Maximum intensity:\\(\cos^2\dfrac{\phi}{2}=1\\)\\(\phi=2n\pi\\) where \\(n=0,1,2,3,\\)....Therefore, \\(I_{max}=4a^2\\)Minmum intensity:\\(\cos^2\dfrac{\phi}{2}=0\\)\\(\phi=(2n+1)\pi\\) where \\(n=0,1,2,3,\\)....Therefore, \\(I_{max}=0\\)

In Young's double slit experiment, the slits are 2mm apart are illuminated by photons of two wavelengths \\({ I }_{ 1 }=12000\mathring { A } \\) and \\({ I }_{ 2 }=10000\mathring { A } \\). at what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Now, from the question we can infer that\\(n_1\lambda_1 = n_2\lambda_2\\)so,\\(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}\\)or \\(\frac{n_1}{n_2} = \frac{10000A}{12000A}\\)thus,we have\\(\frac{n_1}{n_2} = \frac{5}{6}\\)5th and 6th fringes will coincide respectively.the minimum distance is given as\\(X_{min} =\frac{ n_1\lambda_1D}{d}\\)here,\\(n_1 = 5\\)D = 2md = 2mm = 2\\(\times10^{-3}m\\)so,\\(X_{min} = \frac{[5\times12000\times10^{-10}\times2]}{2\times10^{-3}}\\)thus, we get\\(X_{min}\\) = 6mm hence B option is the correct answer

Important Quantities Derivation from YDSE | Definition, Examples, Diagrams

example

Shift of fringes when a slab/lens in introduced in front of a slit

Example:

n

transparent slabs of refractive index

1.5

each, having thickness

1 c m, 2 c m, . . . . . . . . . .

n c m

are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is

1 c m

then find the value of

n

.

Solution:
As the refractive index of all the slabs are

1.5

so those may be a considered as a single slab with the thickness summed up.
Summed up thickness of

n

slabs

Δ t = 1 + 2 + 3 + . . . + n

= \frac{n ( n + 1 )}{2}

Given, for small angle of incidence, shift of object:

s = (1 - \frac{1}{μ}) Δ t

so,

\Rightarrow 1 c m = (1 - \frac{1}{1 . 5}) \frac{n ( n + 1 )}{2}

\Rightarrow 6 = n (n + 1)

\Rightarrow n^{2} + n - 6 = 0

\Rightarrow n^{2} + 3 n - 2 n - 6 = 0

\Rightarrow n (n + 3) - 2 (n + 3) = 0

\Rightarrow (n + 3) (n - 2) = 0

So,

n = - 3

n = 2

As,

n

can not be negative. so

n = 2

example

Define and calculate fringe width and angular fringe width in YDSE

Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. It is given by:

β = \frac{λ D}{d}

Angular fringe width is given by:

tan θ \approx θ = \frac{β}{D} = \frac{λ}{d}

Example: In Young's double slit experiment the two slits are illuminated by light of wavelength

589 0^{\circ} A

and the distance between the fringes obtained on the screen is

0 . 2^{\circ}

. If the whole apparatus is immersed in water then find the angular fringe width. (refractive index of water is

4 / 3

)

Solution:
For interference in YDSE:

d sin θ = n λ

For small

θ

sin θ

can be approximated to

θ

So,

d θ = n λ

or,

θ \propto λ

Now as the set up immersed in water,

λ

will become

\frac{λ}{μ} = \frac{λ}{4 / 3} = \frac{3 λ}{4}

So,

\frac{λ _{1}}{λ _{2}} = \frac{θ _{1}}{θ _{2}}

or,

\frac{λ}{3 λ / 4} = \frac{0 . 2 ^{\circ}}{θ _{2}}

or,

θ_{2} = 0 . 2^{\circ} \times \frac{3}{4} = 0.1 5^{\circ} .

example

Find the ratio of maximum to minimum intensity

Example: In Young's double slit experiment, first slit has width four times the width of the second slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe system.

Solution:

Let intensity from the smaller slit be

I

. So, intensity from the bigger slit will be

4 I

since it is

4

times of smaller slit.
Now,

I_{m a x} = 4 I + I + 2 (4 I^{2})

= 5 I + 2 4 I^{2}

= 5 I + 2 (2 I)

= 9 I

I m i n = 4 I + I - 2 I (4 I)

= 5 I - 2 (4 I^{2})

= I .

So,

\frac{I _{m a x}}{I _{m i n}} = \frac{9 I}{I} = \frac{9}{1} .

Important Quantities Derivation from YDSE

example

Shift of fringes when a slab/lens in introduced in front of a slit

example

Define and calculate fringe width and angular fringe width in YDSE

example

Find the ratio of maximum to minimum intensity

REVISE WITH CONCEPTS

Young's Double Slit Experiment

Quick Summary With Stories

Minimum and Maximum Intensity in Young's Double Slit Experiment

Displacement of Fringes Due to Glass Slab

Path Difference by a Slab and Shifting of Fringes in YDSE

In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

The maximum intensity in Young's double slit experiment is $I_{0}$ . Distance between the slits is $d = 5 λ$ , where $λ$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10 d$ ?

Two coherent monochromatic light beams of intensities $I$ and $4 I$ are superposed. The maximum and minimum possible resulting intensities are :

Derive the expression for the fringe width in a Young's double slit experiment. How will the fringe width change if (i) separation between the slits is increased (ii) screen is moved away from the plane of the slits.

In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

The interference pattern is obtained with two coherent light source of intensity ratio n. In the interference pattern, the ratio $\frac{I _{m a x} - I _{m i n}}{I _{m a x} + I _{m i n}}$ will be

In Young's double slit experiment, the fringe width is found to be $0.4$ mm. If the whole apparatus is immersed in water of refractive index $(4 / 3)$ , without disturbing the geometrical arrangement, what is the new fringe width?

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for the maximum and zero intensity.

example

Shift of fringes when a slab/lens in introduced in front of a slit

example

Define and calculate fringe width and angular fringe width in YDSE

example

Find the ratio of maximum to minimum intensity

REVISE WITH CONCEPTS

Young's Double Slit Experiment

Quick Summary With Stories

Minimum and Maximum Intensity in Young's Double Slit Experiment

Displacement of Fringes Due to Glass Slab

Path Difference by a Slab and Shifting of Fringes in YDSE

In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

The maximum intensity in Young's double slit experiment is I0​. Distance between the slits is d=5λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D=10d?

Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible resulting intensities are :

Derive the expression for the fringe width in a Young's double slit experiment. How will the fringe width change if (i) separation between the slits is increased (ii) screen is moved away from the plane of the slits.

In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

The interference pattern is obtained with two coherent light source of intensity ratio n. In the interference pattern, the ratio Imax​+Imin​Imax​−Imin​​ will be

In Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index (4/3), without disturbing the geometrical arrangement, what is the new fringe width?

Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for the maximum and zero intensity.

The maximum intensity in Young's double slit experiment is $I_{0}$ . Distance between the slits is $d = 5 λ$ , where $λ$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10 d$ ?

Two coherent monochromatic light beams of intensities $I$ and $4 I$ are superposed. The maximum and minimum possible resulting intensities are :

The interference pattern is obtained with two coherent light source of intensity ratio n. In the interference pattern, the ratio $\frac{I _{m a x} - I _{m i n}}{I _{m a x} + I _{m i n}}$ will be

In Young's double slit experiment, the fringe width is found to be $0.4$ mm. If the whole apparatus is immersed in water of refractive index $(4 / 3)$ , without disturbing the geometrical arrangement, what is the new fringe width?