In a Young's double slit experiment, the separation between the slits is 0.15 m m . In the experiment, a source of light of wavelength 589 n m is used and the interference pattern is observed on a screen kept 1.5 m  away. The separation between the successive bright fringes on the screen is:

Option 1 -

4.9 m m

Option 2 -

6.9 m m

Option 3 -

3.9 m m

Option 4 -

5.9 m m

0 5 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 4


    Detailed Solution:

      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

     

     

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

V
Vishal Baghel

In primary rainbow, red colour is at top and violet is at bottom.

Intensity of secondary rainbow is less in comparison to primary rainbow.

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