Figure shows a standard two slit arrangement with slits S1, S2, P1, P2 are the two minima points on either side of P (figure).

At P2 on the screen, there is a hole and behind P2.is a second 2-slit arrangement with slits S3, S4 and a second screen behind them.
(a) There would be no interference pattern on the second screen but it would be lighted
(b) The second screen would be totally dark
(c) There would be a single bright point on the second screen
(d) There would be a regular two slit pattern on the second screen

0 5 Views | Posted 4 months ago
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  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    This is a Multiple Choice Questions as classified in NCERT Exemplar

    Answer- d

    Explanation- According to question, there is a hole at point P2. From Huygen's principle, wave will propagates from the sources S1 and S2. Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P2 (minima). The hole will act as a source of fresh light for the slits S3 and S4.

    Therefore, there will be a regular two slit pattern on the second screen

Similar Questions for you

A
alok kumar singh

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

R
Raj Pandey

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)

P
Payal Gupta

I=I0cos230°

=I0 (32)2=34I0 

V
Vishal Baghel

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

d y D = ( 2 n + 1 ) λ 2   [for Dark fringe at P]

n = 0, for first dark fringe

d y D = λ 2

λ = 2 d D y

= 2 d D × d 2 [ y = d 2 , G i v e n  first dark fringe is observed on the screen directly opposite to one of the slits]

λ = 2 × 0 . 6 × 0 . 6 8 0 0 × 2

λ = 4 5 0 m m

A
alok kumar singh

  Energy     Volume   = M L 2 T - 2 L 3 = M L - 1 T - 2

 The distance between two successive bright fringes is fringe width ( β ) .

β = λ D d = 589 × 10 - 9 × 1.5 0.15 × 10 - 3 = 5.9 m m

 

 

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