Home
Class 12
PHYSICS
In a single slit diffraction experiment ...

In a single slit diffraction experiment first minima for `lambda_1 = 660nm` coincides with first maxima for wavelength `lambda_2`. Calculate the value of `lambda_2`.

Text Solution

Verified by Experts

For minima in diffraction pattern `dsintheta=nlamda`
for first minima `dsintheta_(1)=(1)lamda_(1)impliesintheta_(1)=(lamda_(1))/(d)`
for first maxima `dsintheta_(2)=(3)/(2)lamda_(2)impliessintheta_(2)=(3lamda_(2))/(2d)`
The two will coincide if `theta_(1)=theta_(2)" or "sintheta_(1)=sintheta_(2)`
`therefore(lamda_(1))/(d)=(3lamda_(2))/(2d)implieslamda_(2)=(2)/(3)lamda_(1)=(2)/(3)xx660nm=440nm`
Promotional Banner

Topper's Solved these Questions

  • WAVE OPTICS

    ALLEN |Exercise Example 1|1 Videos
  • WAVE OPTICS

    ALLEN |Exercise Example 2|1 Videos
  • UNIT & DIMENSIONS, BASIC MATHS AND VECTOR

    ALLEN |Exercise Exercise (J-A)|7 Videos

Similar Questions

Explore conceptually related problems

In a single slit diffraction with lambda = 500 nm and a lens of diameter 0.1 mm then width of central maxima, obtain on screen at a distance of l m will be ......

For a given slit, ratio of diffraction angle of fringes of first maxima and first minima is......

In Fraunhoffer diffraction by a single slit, a position where first order minimum is formed by the wavelength of 9000 Å first order maximum is formed due to an unknown wavelength lamda ' is ....

The system of linear equations x+lambday-z=0 lambdax-y-z=0 x+y-lambdaz=0 has a non-trivial solution for : (1) infinitely many values of lambda . (2) exactly one value of lambda . (3) exactly two values of lambda . (4) exactly three values of lambda .

Let S_(1) and S_(2) be the two slits in Young's double-slit experiment. If central maxima is observed at P and angle /_ S_(1) P S_(2) = theta , then fringe width for the light of wavelength lambda will be (assume theta to be a small angle)

What will be the diffraction angle of first order maxima in diffraction obtained due to light of wavelength 55nm and width of slit 0.55 mm ?

Two protons are having same kinetic energy. One proton enters a uniform magnetic field at right angles ot it. Second proton enters a uniform electric field in the direction of field. After some time their de Broglie wavelengths are lambda_1 and lambda_2 then (a) lambda_1 = lambda_2 (b) lambda_1 lt lambda_2 (c) lambda_1 gt lambda_2 (d) some more information is required

Visible light of wavelength 6000 xx 10^(-8) cm fall normally on a single slit and produces diffraction pattern. It is found that the second diffraction minima is at 60^(@) from the centre maxima. If the first minima is produced at theta is close to

In Young's double slit experiment, the two slits act as coherent sources of equal amplitude 'a' and of wavelength lambda . In another experiment with the same setup the two slits are sources of equal amplitude 'a' and wavelength lambda but are incoherent. The ratio of intensities of light at the mid point of the screen in the first case to that in the second case is ....

The potential energy of a partical varies as . U(x) = E_0 for 0 le x le 1 = 0 for x gt 1 for 0 le x le 1 de- Broglie wavelength is lambda_1 and for xgt1 the de-Broglie wavelength is lambda_2 . Total energy of the partical is 2E_0 . find (lambda_1)/(lambda_2).