## wave-optics Question 17

Question: Q. 6. A beam of light consisting of two wavelengths 800 $\mathrm{nm}$ and $600 \mathrm{~nm}$ is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed $1.4 \mathrm{~m}$ away. If the two slits are separated by $0.28 \mathrm{~mm}$, calculate the least distance from the central bright maxima where the bright fringes of the two wavelengths coincide.

A [O.D. I, II, III 2012]

## Show Answer

Solution:

Ans. The two bright fringes will coincide when

$$ m \lambda_{1}=(m+1) \lambda_{2} $$

$$ m \times 800 \times 10^{-9}=(m+1) \times 600 \times 10^{-9} $$

$\therefore \quad m=3$

$$ X_{m}=\frac{m D \lambda_{1}}{d} $$

$X_{m}=\frac{3 \times 1.4 \times 800 \times 10^{-9}}{0.28 \times 10^{-3}} \mathrm{~m}$

$X_{m}=12 \times 10^{-8} \mathrm{~m}$

$1 / 2$

[CBSE Marking Scheme 2012]

## Commonly Made Error

- Many students Couldn’t understand this problem.
- The formula used by a number of students was incorrect.

Q7. Name the phenomenon which is responsible for bending of light around sharp corners of an obstacle. Under what conditions does this phenomenon take place? Give one application of this phenomenon in everyday life.

R [SQP 2014]

Ans. Diffraction.

Diffraction Condition : The size of the obstacle sharpness should be comparable to the wavelength of the light falling. $1+1 / 2$ Application : The finite resolution of our eye. $1 / 2$

[CBSE Marking Scheme 2014]

[AI Q. 8. Write the distinguishing features between a diffraction pattern due to a single slit and the interference fringes produced in Young’s double slit experiment. $\quad$ [OD Comptt. I, II, III 2013]

Ans. Diffraction due to a Single Slit :

(i) It is produced due to different parts of same wavefront.

(ii) Central fringe is twice as wide as other fringes.

(iii) Intensity of fringes decreases as we go to successive maxima away from the centre.

(iv) At an angle $\frac{\lambda}{a}$, first minima is obtained.

Interference Fringe due to Young’s Double Slit :

(i) It is produced due to two different wavefronts.

(ii) Fringe width is of same size.

(iii) Fringes have same intensity.

(iv) At an angle $\frac{\lambda}{a}$, maxima is obtained.