Q.

In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).                  [2008]

1 If d=λ, the screen will contain only one maximum  
2 If λ<d<2λ, at least one more maximum (besides the central maximum) will be observed on the screen  
3 If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase  
4 If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase  

Ans.

(1, 2)

Condition to obtain maxima in Young's double slit experiment is

dsinθ=nλ  where n is an integer.

(1)  When d=λ

       λsinθ=nλ⇒sinθ=n

       When n=0, θ=0

       When n=1, θ=90°

       (This will be a point on the screen which will be at infinity and therefore not practical.)

        Other values of n are invalid as -1≤sinθ≤1.

        Hence, only one maxima when  d=λ.

(2)   When λ<d<2λ

        ⇒λ<nλsinθ<2λ     (∵d=nλsinθ)

         ⇒1<nsinθ<2

          ∴ Possible values of n are 0, +1, -1

           Hence, there is at least one more maxima besides the central maxima.

           As we know,

           Imax=(I1+I2)2  and  Imin=(I1-I2)2

           Initially,  I1=4I  and  I2=I

           ∴ Imax=9I  and  Imin=I

           When I1=I2=I  then Imax=4I  and  Imin=0

           i.e., when the intensities become equal, Imin reduces to zero.