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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference between the interfering waves is $$\lambda$$, ($$\lambda$$ being the wavelength of light used) is I. The intensity at a point where the path difference is $${\lambda \over 4}$$ will be (assume two waves have same amplitude)

A
zero
B
I
C
$${I \over 2}$$
D
$${I \over 4}$$

Explanation

Intensity at any point on the screen is given by

$$I = 4{I_0}{\cos ^2}{\phi \over 2}$$

where, $${I_0}$$ is the intensity of either wave and $$\phi$$ is the phase difference between two waves.

Phase difference $$(\phi ) = {{2\pi } \over \lambda } \times $$ path difference

$$ = {{2\pi } \over \lambda } \times \lambda = 2\pi $$

$$\therefore$$ $$I = 4{I_0}{\cos ^2}\left( {{{2\pi } \over 2}} \right)$$

$$ = 4{I_0}{\cos ^2}(\pi ) = 4{I_0} = I$$

$$ \Rightarrow {I_0} = {I \over 4}$$ ...... (i)

When path difference is $${\lambda \over 4}$$, then

$$\phi = {{2\pi } \over \lambda } \times {\lambda \over 4} = {\pi \over 2}$$

$$\therefore$$ $$I = 4{I_0}{\cos ^2}\left( {{\pi \over 4}} \right) = 2{I_0} = 2 \times {I \over 4} = {I \over 2}$$

ইয়ং-এর দ্বি-ছিদ্র পরীক্ষায় পর্দার উপর একটি বিন্দুতে তীব্রতা I যেখানে উপরিপাতিত তরঙ্গদুটির পথ পার্থক্য $$\lambda$$ ($$\lambda$$ = ব্যবহৃত আলোর তরঙ্গদৈর্ঘ্য) । যে বিন্দুর পথ পার্থক্য $${\lambda \over 4}$$, সেই বিন্দুর তীব্রতা হবে (ধরে নাও, দুটি তরঙ্গের বিস্তার সমান)।

A
শূন্য
B
I
C
$${I \over 2}$$
D
$${I \over 4}$$
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
In Young's double slit experiment, light of wavelength $$\lambda$$ passes through the double slit and forms interference fringes on a screen 1.2 m away. If the difference between 3rd order maximum and 3rd order minimum is 0.18 cm and the slits are 0.02 cm apart, then $$\lambda$$ is
A
1200 nm
B
450 nm
C
600 nm
D
300 nm

Explanation

Given, D = 1.2 m, d = 0.02 cm = 0.02 $$\times$$ 10$$-$$2 m

According to question,

$${{3D\lambda } \over d} - {{5D\lambda } \over {2d}} = 0.18 \times {10^{ - 2}}$$

$$\lambda \left[ {{{3 \times 1.2} \over {0.02 \times {{10}^{ - 2}}}} - {{5 \times 1.2} \over {2 \times 0.02 \times {{10}^{ - 2}}}}} \right] = 0.18 \times {10^{ - 2}}$$

$$ \Rightarrow \lambda \left[ {3 - {5 \over 2}} \right] \times 1.2 = 0.18 \times {10^{ - 2}} \times 0.02 \times {10^{ - 2}}$$

$$ \Rightarrow \lambda = {{0.18 \times {{10}^{ - 4}} \times 0.02 \times 2} \over {1.2}}$$ m

$$ \Rightarrow \lambda = 0.006 \times {10^{ - 4}}$$ m

$$ \Rightarrow \lambda = 600 \times {10^{ - 9}}$$ m

= 600 nm
ইয়ং-এর দ্বি-রেখাছিদ্র পরীক্ষায় $$\lambda $$ তরঙ্গদৈর্ঘ্যের আলাে, ছিদ্র থেকে 1.2 m দূরের পর্দায় ব্যতিচার ঝালর তৈরি করে। যদি ছিদ্র দুটির মধ্যেকার দূরত্ব 0.02 cm হয় এবং তৃতীয় উজ্জ্বল পটি ও তৃতীয় অন্ধকার পটির মধ্যেকার দূরত্ব 0.18 cm হয়ে তবে $$\lambda $$ এর মান হল
A
1200 nm
B
450 nm
C
600 nm
D
300 nm

Explanation

দেওয়া, D = 1.2 m, d = 0.02 cm = 0.02 $$\times$$ 10$$-$$2 m

প্রশ্ন অনুযায়ী,

$${{3D\lambda } \over d} - {{5D\lambda } \over {2d}} = 0.18 \times {10^{ - 2}}$$

$$\lambda \left[ {{{3 \times 1.2} \over {0.02 \times {{10}^{ - 2}}}} - {{5 \times 1.2} \over {2 \times 0.02 \times {{10}^{ - 2}}}}} \right] = 0.18 \times {10^{ - 2}}$$

$$ \Rightarrow \lambda \left[ {3 - {5 \over 2}} \right] \times 1.2 = 0.18 \times {10^{ - 2}} \times 0.02 \times {10^{ - 2}}$$

$$ \Rightarrow \lambda = {{0.18 \times {{10}^{ - 4}} \times 0.02 \times 2} \over {1.2}}$$ m

$$ \Rightarrow \lambda = 0.006 \times {10^{ - 4}}$$ m

$$ \Rightarrow \lambda = 600 \times {10^{ - 9}}$$ m

= 600 nm
3

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
In a Fraunhofer diffraction experiment, a single slit of width 0.5 mm is illuminated by a monochromatic light of wavelength 600 nm. The diffraction pattern is observed on a screen at a distance of 50 cm from the slit. What will be the linear separation of the first order minima?
A
1.0 mm
B
1.1 mm
C
0.6 mm
D
1.2 mm

Explanation

Given, slit width, a = 0.5 mm = 0.5 $$ \times $$ 10-3 m

Wavelength, $$\lambda $$ = 600 nm = 600 $$ \times $$ 10-9 m

Distance of screen, D = 50 cm = 50 $$ \times $$ 10-2 m

$$ \therefore $$ Linear separation = $${{2D\lambda } \over a}$$

= $${{2 \times 50 \times {{10}^{ - 2}} \times 600 \times {{10}^{ - 9}}} \over {0.5 \times {{10}^{ - 3}}}}$$

= 12 $$ \times $$ 10-4 m

= 1.2 $$ \times $$ 10-3 m = 1.2 mm

ফ্রাউনহোফার-এর অপবর্তন পরীক্ষায় 0.5 mm বেধের একটি ছিদ্রকে 600 nm তরঙ্গ দৈর্ঘ্যের একবর্ণী আলো দ্বারা প্রভাসিত করা হল এবং 50 cm দূরে একটি পর্দার উপর অপবর্তন সজ্জা তৈরি হল। সেক্ষেত্রে দুটি প্রথম কৃষ্ণপটির মধ্যে রৈখিক দূরত্ব কত ?

A
1.0 mm
B
1.1 mm
C
0.6 mm
D
1.2 mm

Explanation

d = 0.5 mm ; D = 50 cm ; $$\lambda$$ = 600 nm

কেন্দ্রীয় চরমপটির প্রস্থ = $${{2\lambda D} \over d}$$

$$ = {{2 \times 600 \times 50 \times {{10}^{ - 7}}} \over {0.5 \times {{10}^{ - 6}}}}$$

= 1.2 nm

$$\Rightarrow$$ Option (d) সঠিক।

4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
The intensity of light emerging from one of the slits in a Young's double slit experiment is found to be 1.5 times the intensity of light emerging from the other slit. What will be the approximate ratio of intensity of an interference maximum to that of an interference minimum?
A
2.25
B
98
C
5
D
9.9

Explanation

Let intensity from one slit = I1
Intensity from other slit = I2
According to question,
I1 = 1.5I2
$$ \Rightarrow $$ $${{{I_1}} \over {{I_2}}} = 1.5 = {3 \over 2}$$
$$ \therefore $$ Ratio of intensity, $${{{I_{\max }}} \over {{I_{\min }}}} = {{{{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)}^2}} \over {{{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)}^2}}}$$ = $${\left( {{{\sqrt 3 + \sqrt 2 } \over {\sqrt 3 - \sqrt 2 }}} \right)^2}$$
= $${\left( {{{1.732 + 1.414} \over {1.732 - 1.414}}} \right)^2}$$
= $${\left( {{{3.146} \over {0.318}}} \right)^2}$$ = $${\left( {9.89} \right)^2}$$ $$ \simeq $$ 98

ইয়ং-এর দ্বি-ছিদ্র পরীক্ষায় দেখা গেল, এক ছিদ্র থাকে নির্গত আলোর তিব্রতা অন্য ছিদ্র থেকে নির্গত আলোর তিব্রতার 1.5 গুণ। সেক্ষেত্রে সর্বোচ্চ ও সর্বনিম্ন ব্যতিচারের আলোর তিব্রতার অনুপাত কত ?

A
2.25
B
98
C
5
D
9.9

Explanation

শর্তানুসারে, I1 = 1.5 I2

$$ \Rightarrow {{{I_1}} \over {{I_2}}} = {3 \over 2}$$

$${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} = {\left( {{{\sqrt 3 + \sqrt 2 } \over {\sqrt 3 - \sqrt 2 }}} \right)^2} = 98$$

$$\Rightarrow$$ Option (b) সঠিক।

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