Practice free Wave Optics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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
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 $ . With the light of wavelength $ 5000 A^\circ $ , If experiment were carried out under water of a n = 4 /3 the fringe width would be
When the experiment is carried out in a medium with refractive index \( n \), the wavelength of light changes to \( \lambda' = \frac{\lambda}{n} \). Therefore, the new fringe width \( \beta' \) is given by \( \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} \). Thus, the fringe width in water (where \( n = \frac{4}{3} \)) will be \( \frac{3}{4} \) times the fringe width in air.
In a fraunhofer diffraction by single slit of width d with incident light of wavelength $ 5500 A ^\circ $ the first minimum is observed at angle of $30 ^\circ $ . The first secondary maximum is observed at an angle $ \theta $ =
The phenomenon of polarisation of electromagnetic waves proves that the electromagnetic waves are
The phenomenon of polarization is characteristic of transverse waves, not longitudinal waves. In polarization, the oscillations of the electromagnetic wave are restricted to a particular direction perpendicular to the direction of wave propagation. This can only happen if the waves are transverse. Therefore, polarization proves that electromagnetic waves are transverse in nature.
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