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Class 10 Class 12

A double-slit is illuminated by light of wavelength 6000A. The slits are 0.1 cm apart and the screen is placed 1 m away. Calculate (i) angular position of 10 th maximum in radian (ii) separation of two adjacent minima.

Given ,
Wavelength of the light source,

λ

= 6000

\overset{o}{A}

Distance between the slits, d = 0.1 cm = 0.1

\times 10^{- 2} m

Distance between the screen and the slit, D = 1 m

(i) Angular position of nth maximum is given by,

θ_{n} = \frac{x_{n}}{D} = \frac{nDλ}{dD} = \frac{nλ}{d} ∴ χ_{n} = \frac{n D λ}{d}

Angular position of 10th maximum,

θ_{10} = \frac{10 \times 6000 \times 10^{- 10}}{0.1 \times 10^{- 2}} = 6 \times 10^{- 3} radian

(ii) Separation between two adjacent minima

i.e., fringe width,

ω = \frac{Dλ}{d}

\Rightarrow

ω = \frac{1 \times 6000 \times 10^{- 10}}{10^{- 3}}

= 6 \times 10^{- 4} m = 0.6 mm .

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How would the angular separation of interference fringes in Young’s double-slit experiment change when the distance of separation between slits and screen is doubled?

In young's double slit experiment,
Angular separation is given by,
$θ = \frac{β}{D} = \frac{λ}{d} .$ $[∵ fringe width β = \frac{λD}{d}]$

Angular seperation does not depend on D i.e., the distance of separation between slits and screen. Therefore, $θ$ remains unaffected.

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What is diffraction of light? Draw a graph showing the variation of intensity with angle in a single slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern.
How would the diffraction pattern of a single slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?

Diffraction of light: Phenomenon of bending of light around the comers of an obstacle or aperture is called diffraction.
The intensity distribution wave for diffraction is shown in the diagram below.
$Diffraction of light: Phenomenon of bending of light around the comer$

In interference, by 2 slits all bright fringes are of same intensity. In diffraction, the intensity of bright fringes decreases with the increase in distance from the central bright fringe.
(i) The diffraction pattern becomes narrower if the width of the slit is decreased.
(ii) A coloured diffraction pattern is obtained if monochromatic source is replaced by white light source. The central band is white therefore, the red fringe is wider than the violet fringe etc.

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Two convex lens of same focal length but their aperture and focal lengths 5 cm and 10 cm are used as object lens in two astronomical telescope.
(a) What will be the ratio of their
(i) resolving power
(ii) magnifying power
(b) Compare the intensity of images formed in these cases.

Given, two convex lens of same focal length.
Aperture of the first lens, D₁ = 5 cm
Aperture of second lens, D₂ = 10 cm

(i) Ratio of resolving power is,

\frac{\frac{D_{1}}{1.22 λ}}{\frac{D_{2}}{1.22 λ}} i . e ., \frac{D_{1}}{D_{2}} = \frac{5}{10} = \frac{1}{2}

(ii) Ratio of magnifying power is given by,

\frac{{(\frac{f_{0}}{f_{e}})}_{1}}{{(\frac{f_{0}}{f_{e}})}_{2}} = \frac{{(f_{0})}_{1}}{(f_{0})_{2}} = \frac{5}{10} = \frac{1}{2}

(b) Intensity of image is directly proportional to the size of aperture hence, it will also have a ratio of

\frac{1}{2} .

200 Views

In a Young's experiment, the width of the fringes obtained with light of wavelength 6000A is 2.0 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33?

Given,

Width of fringes, ω = 2 \times 10^{- 3} m Wavelength of light used, λ = 6000 Å = 6 \times 10^{- 7} m

The formula for fringe width is,

ω = \frac{Dλ}{d}

\Rightarrow

\frac{D}{d} = \frac{ω}{λ} = \frac{2 \times 10^{- 3}}{6 \times 10^{- 7}} = \frac{1}{3} \times 10^{4}

When the apparatus is immersed in liquid

Wavelength,

λ' = \frac{λ}{μ} = \frac{6 \times 10^{- 7}}{1.33} m

Fringe width,

ω' = \frac{D}{d} λ'

\Rightarrow

ω' = \frac{1}{3} \times 10^{4} \times \frac{6 \times 10^{- 7}}{1.33} m

= 1.5 \times 10^{- 3} m = 1.5 mm .

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