Wave Optics - Result Question 10
10. In Young’s double slit experiment the separation $d$ between the slits is $2 mm$, the wavelength $\lambda$ of the light used is $5896 \AA$ and distance $D$ between the screen and slits is $100 cm$. It is found that the angular width of the fringes is $0.20^{\circ}$. To increase the fringe angular width to $0.21^{\circ}$ (with same $\lambda$ and D) the separation between the slits needs to be changed to
(a) $1.8 mm$
(b) $1.9 mm$
(c) $1.7 mm$
(d) $2.1 mm$
[2018]
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Answer:
Correct Answer: 10. (b)
Solution:
- (b) Angular width $=\frac{\lambda}{d}$
$ \begin{aligned} & \text{ So, } 0.20^{\circ}=\frac{\lambda}{2 mm} \\ & \Rightarrow \quad \lambda=0.20^{\circ} \times 2 \\ & \text{ Again, } 0.21^{\circ}=\frac{\lambda}{d} \end{aligned} $
Now putting the value of $\lambda$
$ \begin{aligned} d & =\frac{0.20^{\circ} \times 2 mm}{0.21^{\circ}} \\ \therefore \quad d & =1.9 mm \end{aligned} $
