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nth bright fringe of red light $ ( \lambda = 7500 A ^\circ ) $. Coincides with
(n+1) th bright fringe of green light $ ( \lambda_2 = 6000 ^\circ ) $. The value of n = _____
$ use \;n \lambda_1 = ( n+1) { \lambda \over 2 } $ $ \therefore n = 4 $
Which of the following will undergo maximum diffration ?
Diffraction is more pronounced for waves with larger wavelengths. Among the given options, radio waves have the longest wavelength. Therefore, they will undergo the maximum diffraction.
A Slit of width $ 12 \times 10^ {-17 } $ is illuminated by light of wavelenth $ 6000 A ^ \circ $ . The angular width of the central maxima is appoximately__ .
The angular width of the central maximum in a single-slit diffraction pattern is given by $2 heta = rac{2 ext{λ}}{a}$, where $ ext{λ}$ is the wavelength and $a$ is the slit width. Substituting $ ext{λ} = 6000 ext{Å}$ and $a = 12 imes 10^{-7} ext{cm}$, we get an angular width of approximately $60^ ext{°}$.
The distance between the first and sixth minima in the diffraction pattern of a single slit, it is 0.5 mm. The screen is 0.5 m away from the Slit. If the wavelength of light is $ 5000 A ^ \circ $ , then the width of the slit will be_______ mm
_ _ _ _ change in the polarization phynomina of ligst ?
Polarization refers to the orientation of the oscillations of light waves. One of the key effects of polarization is the change in the intensity of light. When light is polarized, its intensity can vary depending on the angle and the method of polarization. Therefore, the correct answer is that intensity changes in the polarization phenomenon of light.
In yong's double slit experiment the phase diffrence is constant between two sources is $ \pi /2 $. The intensity at a point equi distant from the slits in terms of max. intensity $I_o$ is........
The two coherent sources of intensity β produce interference. The fringe visibility will be_
Fringe visibility (V) in an interference pattern is given by the formula: V = (I_max - I_min) / (I_max + I_min). For two coherent sources of equal intensity β, the fringe visibility is calculated using V = (2√β) / (1 + β). Hence, the correct option is $\frac{2\sqrt{\beta}}{1+\beta}$.
Light of wave–length $ \lambda $ is incident on a slit of width d. The resulting diffraction pattern is observed on a screen placed at a distance D. The linear width of the principal maximum is equal to the width of the slit, then D = ______.
The linear width of the principal maximum in a single-slit diffraction pattern is given by $2\lambda D / d$. If this width is equal to the width of the slit (d), then we can set up the equation $2\lambda D / d = d$. Solving for D, we get $D = \frac{d^2}{2\lambda}$. Hence, the correct option is $\frac{d^2}{2\lambda}$.
A polariser is used for
A polarizer is a device that converts unpolarized light into polarized light by allowing only light waves oscillating in a particular direction to pass through. Hence, the correct option is 'Produced polarised light'.
Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given $ x = n \lambda { D \over d } $ and the positon of dark fringes is given by $ x = (2n -1 ) { \lambda \over 2 } { D \over d } $ where n = 1,2,3........... for first, second, third bright / dark fringe. The center of the fringe pattern is bright (for n = 0). The width of each briht/dark fringe is $ \lambda = 5000 A ^\circ $ . If light of wavelength $6000 A^\circ $ be used in the above experiment the fringe width would be ........mm
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.