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Standard XII

Physics

Maxima & Minima in YDSE

In YDSE dista...

Question

In YDSE distance between slits and screen is $1.5 m$ . When light of wavelength $500 nm$ is used then $2^{n d}$ bright fringe is obtained on screen at a distance of $10 mm$ from central bright fringe. What will be shift in position of $2^{n d}$ bright fringe, if light of wavelength $550 nm$ is used ?

2 mm

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1 mm

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1.5 mm

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1.1 mm

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Solution

The correct option is B $1 mm$
Given
$D = 1.5 mm λ_{1} = 500 nm, y_{1} = 10 mm,$
$λ_{2} = 550 nm, y_{2} = ?$
As $y \propto λ$
$\Rightarrow \frac{y_{1}}{y_{2}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{10}{y_{2}} = \frac{500}{550}$
$\Rightarrow y_{2} = 11 mm \Rightarrow y_{2} - y_{1} = 1 mm$

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In YDSE distance between slits and screen is 1.5 m. When light of wavelength 500 nm is used then 2nd bright fringe is obtained on screen at a distance of 10 mm from central bright fringe. What will be shift in position of 2nd bright fringe, if light of wavelength 550 nm is used ?

The correct option is B 1 mm Given D=1.5 mm λ1=500 nm, y1=10 mm, λ2=550 nm, y2=? As y∝λ ⇒y1y2=λ1λ2⇒10y2=500550 ⇒y2=11 mm⇒y2−y1=1 mm

The correct option is B $1 mm$
Given
$D = 1.5 mm λ_{1} = 500 nm, y_{1} = 10 mm,$
$λ_{2} = 550 nm, y_{2} = ?$
As $y \propto λ$
$\Rightarrow \frac{y_{1}}{y_{2}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{10}{y_{2}} = \frac{500}{550}$
$\Rightarrow y_{2} = 11 mm \Rightarrow y_{2} - y_{1} = 1 mm$