Home
Class 12
PHYSICS
In the fraunhaufer differaction from a s...

In the fraunhaufer differaction from a single slit illuminated by polychromatic light, the first minimum with wavelength `lamda_(1)` is found to be coincident with the third minimum at `lamda_(2)`. Then find the value of `(lamda_(1))/(lamda_(2))`

Text Solution

Verified by Experts

The correct Answer is:
3

NA
Promotional Banner

Similar Questions

Explore conceptually related problems

In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength lambda_1 is found to be coincident with third maximum at lambda_2 . So

In a single slit diffraction pattern, the first minima for a wavelength lamda_(1)=6000 Å coincides with the first maxima for a wavelength lamda_(2) . Then the value of lamda_(2) is

In a biprism experiment, if the 6th bright bannd with a wavelength lamda_(1) coincides with 7th dark band with a wavelength lamda_(2) , then the ratio of the wavelengths ((lamda_(1))/(lamda_(2))) is

What should be the slit width to obtain pronounced diffraction with a single slit illuminated by light of wavelength lamda ?

In a single slit diffraction experiment the first minimum for red light of wavelength 6600Å coincides with the first maximum for other light of wavelength lamda . The value of lamda is

In a single slit diffraction experiment, first minimum for a light of wavelength 540 nm coincides with the first maximum of another wavelength lamda' . Then lamda' is

For Balmer series, wavelength of first line is lamda_(1) and for Brackett series, wavelength of first line is lamda_(2) then (lamda_(1))/(lamda_(2))

In the diffraction pattern due to a single slit of width 'd' with incident light of wavelength 'lamda', at an angle of diffraction 'theta', the condition for first minimum is

A slit of width a is illuminiated by white light. The first diffraction minimum for light of lamda=6500 Å is formed at theta=30^(@) , then the width (a) of the slit is