SATHEE: Optics 6 Question 2

Optics 6 Question 2

3. In a Young’s double slit experiment, the ratio of the slit’s width is $4 : 1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

(a) $4 : 1$

(b) $25 : 9$

(d) $(\sqrt{3} + 1)^{4} : 16$

(2019 Main, 10 April II)

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Solution:

Key Idea Amplitude of light is directly proportional to area through which light is passing.

For same length of slits,

$amplitude \propto (width)^{1 / 2}$

Also,

$intensity \propto (amplitude)^{2}$

In YDSE, ratio of intensities of maxima and minima is given by

$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$

where, $I_{1}$ and $I_{2}$ are the intensities obtained from two slits, respectively.

$\Rightarrow \frac{I_{max}}{I_{min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}}$

where, $a_{1}$ and $a_{2}$ are light amplitudes from slits 1 and 2 , respectively.

$\Rightarrow \frac{I_{max}}{I_{min}} = \frac{{(\sqrt{W_{1}} + \sqrt{W_{2}})}^{2}}{{(\sqrt{W_{1}} - \sqrt{W_{2}})}^{2}}$

where, $W_{1}$ and $W_{2}$ are the widths of slits, respectively.

Here, $\frac{W_{1}}{W_{2}} = {\frac{a_{1}}{a_{2}}}^{2} = \frac{4}{1} \Rightarrow \sqrt{\frac{W_{1}}{W_{2}}} = \frac{2}{1}$

So, we have

$\begin{aligned} \frac{I_{max}}{I_{min}} = \frac{\sqrt{W_{1}} + \sqrt{W_{2}}}{\sqrt{W_{1}} - \sqrt{W_{2}}} & = \frac{\sqrt{\frac{W_{1}}{W_{2}}} + 1}{\sqrt{\frac{W_{1}}{W_{2}}} - 1} \\ = {\frac{2 + 1}{2 - 1}}^{2} = 9 : 1 \end{aligned}$

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Optics 6 Question 2

3. In a Young’s double slit experiment, the ratio of the slit’s width is $4 : 1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics 6 Question 2

3. In a Young’s double slit experiment, the ratio of the slit’s width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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3. In a Young’s double slit experiment, the ratio of the slit’s width is $4 : 1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be