Solution:

Ans. (a) Definition of wavefront Verification of laws of reflection

$1 / 2$

(b) Explanation of the effect on the size and intensity of central maxima

$1+1$

(c) Explanation of the bright spot in the shadow of the obstacle

(a) The wavefront may be defined as a surface of constant phase.

[Alternatively : The wave front is the locii of all points that are in the same phase]

Let speed of the wave in the medium be ’ $v$ '

Let the time taken by the wave front, to advance from point $B$ to point $C$ be ’ $\tau$ '

$$ \text { Hence } \quad B C=v \tau $$

Let $C E$ represent the reflected wave front

Distance $\quad A E=v \tau=B C$

$\triangle A E C$ and $\triangle A B C$ are congruent

$$ \angle B A C=\angle E C A $$

$$ \Rightarrow \quad \angle i=\angle r $$

(b) Size of central maxima reduces to half,

$\left(\therefore\right.$ Size of central maxima $\left.=2 \frac{2 \lambda D}{\alpha}\right)$

Intensity increases.

This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

$1 / 2$

(Also accept if the student just writes that the intensity becomes four fold)

(c) This is because of diffraction of light.

[Alternatively : Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.]

[Alternatively : There is a maxima, at the centre of the obstacle, in the diffraction pattern produced by it.]

[CBSE Marking Scheme 2018]

AI Q. 6. (i) Derive an expression for path difference in Young’s double slit experiment and obtain the condition for constructive and destructive interference at a point on the screen.

(ii) The intensity at the central maxima in Young’s double slit experiment is $I_{0}$. Find out the intensity at a point where the path difference is $\lambda / 6, \lambda / 4$ and $\lambda / 3$.

U] [OD North 2016]

Ans. (i) Try yourself, Similar to Q. 2 (i) LAT Questions.

(ii) Try yourself, Similar to Q. 2 SAT Questions-I