मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

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. - Physics

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

Given:-

Refractive index of the plate is μ.

The thickness of the plate is t.

Wavelength of the light is λ.

(a)

When the plate is placed in front of the slit, then the optical path difference is given by \[\left( \mu - 1 \right)t\]

(b) For zero intensity at the centre of the fringe pattern, there should be distractive interference at the centre.

So, the optical path difference should be = \[\frac{\lambda}{2}\]

\[i . e .   \left( \mu - 1 \right)  t = \frac{\lambda}{2}\]

\[ \Rightarrow t = \frac{\lambda}{2  \left( \mu - 1 \right)}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 17: Light Waves - Exercise [पृष्ठ ३८१]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 17 Light Waves
Exercise | Q 13 | पृष्ठ ३८१

संबंधित प्रश्‍न

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?


Using analytical method for interference bands, obtain an expression for path difference between two light waves.


A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is a distance of 2.5 mm away from the centre. Find the width of the slit.


A Young's double slit experiment is performed with white light.

(a) The central fringe will be white.

(b) There will not be a completely dark fringe.

(c) The fringe next to the central will be red.

(d) The fringe next to the central will be violet.


In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?


A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?


A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta =  \sin^{- 1}   \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.


White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (see the following figure). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?


In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.


In Young’s double-slit experiment, show that: 

`beta = (lambda "D")/"d"` where the terms have their usual meaning.


In Young’s double slit experiment, what should be the phase difference between the two overlapping waves to obtain 5th dark band/fringe on the screen?


In Young's double slit experiment the slits are 0.589 mm apart and the interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 7.5 mm from the dark fringe which is second from the center of the fringe pattern. Find the wavelength of the light used.


Young's double slit experiment is made in a liquid. The 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately


How will the interference pattern in Young's double-slit experiment be affected if the phase difference between the light waves emanating from the two slits S1 and S2 changes from 0 to π and remains constant?


In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate

  1. the fringe width
  2. the distance of (a) third minimum and (b) fifth maximum, from the central maximum.

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit- width. Find the ratio of the maximum to the minimum intensity in the interference pattern.


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 :


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×