Optics 6 Question 52
54. In Young’s experiment, the source is red light of wavelength $7 \times 10^{-7} m$. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by $10^{-3} m$ to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength $5 \times 10^{-7} m$, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength.
(1997C, 5M)
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Answer:
Correct Answer: 54. $7 \times 10^{-6} m, 1.6,-5.71 \times 10^{-5} m$
Solution:
- (a) Path difference due to the glass slab,
$$ \Delta x=(\mu-1) t=(1.5-1) t=0.5 t $$
Due to this slab, 5 red fringes have been shifted upwards.
Therefore,
$$ \begin{aligned} \Delta x & =5 \lambda _{\text {red }} \\ \text { or } \quad 0.5 t & =(5)\left(7 \times 10^{-7} m\right) \end{aligned} $$
$\therefore t=$ thickness of glass slab $=7 \times 10^{-6} m$
(b) Let $\mu^{\prime}$ be the refractive index for green light then
$$ \Delta x^{\prime}=\left(\mu^{\prime}-1\right) t $$
Now the shifting is of 6 fringes of red light. Therefore,
$$ \begin{array}{rlrl} \Delta x^{\prime} & =6 \lambda _{\text {red }} \\ & \therefore & \left(\mu^{\prime}-1\right) t & =6 \lambda _{\text {red }} \\ & \therefore & \left(\mu^{\prime}-1\right) & =\frac{(6)\left(7 \times 10^{-7}\right)}{7 \times 10^{-6}}=0.6 \\ & \therefore & \mu^{\prime} & =1.6 \end{array} $$
(c) In part (a), shifting of 5 bright fringes was equal to $10^{-3} m$. Which implies that
$$ \begin{aligned} & 5 \omega _{\text {red }}=10^{-3} m \quad(\text { Here, } \omega=\text { Fringe width }) \\ & \therefore \quad \omega _{\text {red }}=\frac{10^{-3}}{5} m=0.2 \times 10^{-3} m \\ & \text { Now since } \omega=\frac{\lambda D}{d} \\ & \text { or } \quad \omega \propto \lambda \\ & \therefore \quad \frac{\omega _{\text {green }}}{\omega _{\text {red }}}=\frac{\lambda _{\text {green }}}{\lambda _{\text {red }}} \\ & \therefore \quad \omega _{\text {green }}=\omega _{\text {red }} \frac{\lambda _{\text {green }}}{\lambda _{\text {red }}}=\left(0.2 \times 10^{-3}\right) \frac{5 \times 10^{-7}}{7 \times 10^{-7}} \\ & \omega _{\text {green }}=0.143 \times 10^{-3} m \\ & \therefore \Delta \omega=\omega _{\text {green }}-\omega _{\text {red }}=(0.143-0.2) \times 10^{-3} m \\ & \Delta \omega=-5.71 \times 10^{-5} m \end{aligned} $$
