Advertisements
Advertisements
प्रश्न
Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be placed between them so that the intensity of polarized light transmitted by polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A?
Advertisements
उत्तर
Let θAC, angle between the transmission axis of Polaroid A and Polaroid C.
θCB, angle between the transmission axis of Polaroid C and Polaroid B.
Then,
θAC + θCB = 180° (As, Polaroid ‘A’ and ‘B’ are kept in crossed position.)
Or, θAC = 180° − θCB ... (1)
Intensity of unpolarized light = I0
I1, I2 and I3 are the intensities of light on passing through the A, B and C polarises respectively.
Now,
`I_1 = 1/2I_0 .....(2)`
`I_2 =I_1cos^2theta_(AC)`
=`1/2I_0cos^2theta_(AC)`
`= 1/2I_0 cos^2(180° -theta_(CB))` (from equation (1))
`I_2=1/2I_0sin^2theta_(CB)......... (3)`
and, `I_3 =I_2cos^2theta_(CB)`
`I_3=(1/2I_0sin^2theta_(CB)) cos^2theta_(CB)`
or,`I_3 = 1/2I_0sin_2theta_(CB)cos^2theta_(CB)`
As,given, `I_3=1/8 I_0`
Therefore,`1/8I_0 =1/2I_0 xx 1/4(sin2theta_(CB))^2`
`or, sin 2 theta_(CB =1)`
`or, 2theta_(CB) =90°`
`or, theta_(CB) =45°`
Thus, Polaroid ‘C’ must be placed at angle 45° with Polaroid ‘B’.
संबंधित प्रश्न
What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?
In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is `λ/3`.
In a Young’s 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.
A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is \[\lambda.\] Neglect any absorption of light in the plate.
A double slit S1 − S2 is illuminated by a coherent light of wavelength \[\lambda.\] The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (see the following figure). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to

(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit".
Justify the above statement through a relevant mathematical expression.
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
