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In Young's experiment, light of wavelength 4000 Å is used to produce bright fringes of width 0.6 mm, at a distance of 2 meters. If the whole apparatus is dipped in a liquid of refractive index 1.5, then fringe width will be

0.2 mm

0.3 mm

0.4 mm

1.2 mm

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Explanation

$β_{m e d i u m} = \frac{β_{a i r}}{μ} = \frac{0 .6}{1 .5} = 0 .4 m m$ .

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In Young's double-slit experiment, the phase difference between the light waves reaching the third bright fringe from the central fringe will be (λ =6000 Å )

Zero

2 $π$

4 $π$

6 $π$

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Explanation

$∵ n = 3, ∴ 2 n π = 2 \times 3 π = 6 π$

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In two separate set - ups of the Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1 : 2 are used. If the ratio of the slit separation in the two cases is 2 : 1, the ratio of the distances between the plane of the slits and the screen in the two set - ups is

4 : 1

1 : 1

1 : 4

2 : 1

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Explanation

As $β = \frac{D λ}{d} \Rightarrow \frac{β_{1}}{β_{2}} = (\frac{D_{1}}{D_{2}}) (\frac{λ_{1}}{λ_{2}}) (\frac{d_{2}}{d_{1}})$

$\Rightarrow 1 = (\frac{D_{1}}{D_{2}}) \times (\frac{1}{2}) \times (\frac{1}{2}) \Rightarrow \frac{D_{1}}{D_{2}} = \frac{4}{1}$

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In a biprism experiment, by using light of wavelength 5000 Å, 5 mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is

1.0 mm

0.1 mm

0.05 mm

0.01 mm

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Explanation

$d = \frac{D λ}{β} = \frac{1 \times 5 \times 10^{- 7}}{5 \times 10^{- 3}} = 10^{- 4} m = 0 .1 m m$ .

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The slits in a Young's double-slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringes is I₀. If one of the slits is closed, the intensity at this point will be

I₀

I₀/ 4

I₀/ 2

4I₀

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Explanation

If intensity of each wave is I, then initially at central position I₀ = 4I. when one of the slit is covered then intensity at central position will be I only i.e., $\frac{I_{0}}{4}$ .

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A thin mica sheet of thickness 2×10^–6 m and refractive index (μ = 1.5) is introduced in the path of the first wave. The wavelength of the wave used is 5000 Å. The central bright maximum will shift

2 fringes upward

2 fringes downward

10 fringes upward

None of these

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Explanation

Shift $= \frac{β}{λ} (μ - 1) t = \frac{β}{(5000 \times 10^{- 10})} (1 .5) \times 2 \times 10^{- 6} = 2 β$

i.e., 2 fringes upwards.

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In Young's double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4358 Å is used in place of sodium light, then number of fringes seen will be

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Explanation

$n_{1} λ_{1} = n_{2} λ_{2}$

$\Rightarrow 62 \times 5893 = n_{2} \times 4358$

$\Rightarrow n_{2} = 84 .$

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In Young's double slit experiment, angular width of fringes is 0.20^o for sodium light of wavelength 5890 Å. If complete system is dipped in water, then angular width of fringes becomes

0.11^o

0.15^o

0.22^o

0.30^o

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Explanation

Angular fringe width $θ = \frac{λ}{d} \Rightarrow θ \propto λ$

$λ_{w} = \frac{λ_{a}}{μ_{w}}$

So $θ_{w} = \frac{θ_{air}}{μ_{w}}$ $= \frac{0 .20}{\frac{4}{3}} = 0 .15 °$

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In Young's double slit experiment, the distance between the two slits is 0.1 mm and the wavelength of light used is 4×10^–7 m. If the width of the fringe on the screen is 4 mm, the distance between screen and slit is

0.1 mm

1 cm

0.1 cm

1 m

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Explanation

$β = \frac{λ D}{d} \Rightarrow (4 \times 10^{- 3}) = \frac{4 \times 10^{- 7} \times D}{0 .1 \times 10^{- 3}} \Rightarrow D = 1 m$

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In Young's double slit experiment using sodium light (λ = 5898 Å), 92 fringes are seen. If given colour (λ = 5461 Å) is used, how many fringes will be seen

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Explanation

$(n_{1} λ_{1} = n_{2} λ_{2})$

$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} \Rightarrow \frac{n_{1}}{92} = \frac{5898}{5461} \Rightarrow n_{1} = 99$

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