In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5×10−2 m towards the slits, the change in fringe width is 3×10−5 m. If separation between the slits is 10−3 m, the wavelength of light used is

β=λDd⇒β∝D⇒β1β2=D1D2⇒β1−β2β2=D1−D2D2⇒ΔβΔD=β2D2=λ2d2=λ2=3×10−55×10−2×10−3=6×10−7m=6000Å

In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5×10−2 m towards the slits, the change in fringe width is 3×10−5 m. If separation between the slits is 10−3 m, the wavelength of light used is

β=λDd⇒β∝D⇒β1β2=D1D2⇒β1−β2β2=D1−D2D2⇒ΔβΔD=β2D2=λ2d2=λ2=3×10−55×10−2×10−3=6×10−7m=6000Å

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In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by $5 \times 10^{- 2} m$ towards the slits, the change in fringe width is $3 \times 10^{- 5} m .$ If separation between the slits is $10^{- 3} m$ , the wavelength of light used is

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$4500 Å$

$6000 Å$

$3000 Å$

$5000 Å$

answer is A.

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Detailed Solution

$β = \frac{λD}{d} \Rightarrow β \propto D$

$\Rightarrow \frac{β_{1}}{β_{2}} = \frac{D_{1}}{D_{2}} \Rightarrow \frac{β_{1} - β_{2}}{β_{2}} = \frac{D_{1} - D_{2}}{D_{2}} \Rightarrow \frac{Δβ}{ΔD} = \frac{β_{2}}{D_{2}} = \frac{λ_{2}}{d_{2}}$

$= λ_{2} = \frac{3 \times 10^{- 5}}{5 \times 10^{- 2}} \times 10^{- 3} = 6 \times 10^{- 7} m = 6000 Å$

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In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5×10−2 m towards the slits, the change in fringe width is 3×10−5 m. If separation between the slits is 10−3 m, the wavelength of light used is

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