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# What will be the width of the slit if the distance between the first and third minima in the diffraction pattern is $3mm$? The distance between screen and single slit is $50cm$ which is illuminated with light of wavelength $6000\dot A$.A) $0.1mm$B) $0.2mm$C) $0.3mm$D) $0.4mm$

Last updated date: 29th May 2024
Total views: 47.4k
Views today: 0.47k
Verified
47.4k+ views
-Hint:- Convert all the given values to their S.I unit.
Now, evaluate the distance between the first and third minima.
Next, we can now calculate the width of slit by the expression –
$x = \dfrac{{\lambda D}}{d}$
where, $x$ is the distance between first and third minima,
$\lambda$ is the wavelength of light,
$D$ is the distance between screen and slit and
$d$ is the width of slit.

Complete Step by Step Solution:-
Let the distance between first and third minima be $x$ and first minima be ${x_1}$ and third minima be ${x_3}$. Therefore,
$x = {x_3} - {x_1} \\ x = 3mm \\ x = 3 \times {10^{ - 3}}m \\$
Let the distance between the screen and single slit be $D$. So, according to the question it is given that –
$D = 50cm \\ D = 0.5m \\$
Let the width of slit be $d$ and wavelength of light be $\lambda$.
According to the question, it is given that
$\lambda = 6000\dot A \\ \lambda = 6000 \times {10^{ - 10}}m \\$
The position of ${n^{th}}$ minima in the diffraction pattern is given by –
${x_n} = n\dfrac{{D\lambda }}{d}$
The above can also be rewritten by transposition as –
$d = n\dfrac{{D\lambda }}{x} \cdots (1)$
Putting the values of $x,n,D$ and $\lambda$ in their respective places in equation $(1)$
$d = (3 - 1)\dfrac{{0.5 \times 6000 \times {{10}^{ - 10}}}}{{3 \times {{10}^{ - 3}}}} \\ d = \dfrac{{2 \times 0.5 \times 6 \times {{10}^3} \times {{10}^{ - 10}}}}{{3 \times {{10}^{ - 3}}}} \\$
By further solving, we get
$d = \dfrac{{6 \times {{10}^{ - 4}}}}{3} \\ d = 2 \times {10^{ - 4}}m \\$
Now, converting metre to millimetre, we get
$d = 0.2mm$

Therefore, the correct answer is option (B).

Note:- Diffraction is a specialized case of scattering of light in which an object with regularly repeating features produces an orderly diffraction of light in a diffraction pattern. The diffraction plays an important role in limiting the resolving power of any optical instrument.
The text contains a physics problem that involves a mathematical calculation to determine the width of a slit based on given parameters. The problem is followed by a detailed step-by-step solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# What will be the width of the slit if the distance between the first and third minima in the diffraction pattern is $3mm$? The distance between screen and single slit is $50cm$ which is illuminated with light of wavelength $6000\dot A$.A) $0.1mm$B) $0.2mm$C) $0.3mm$D) $0.4mm$

Last updated date: 29th May 2024
Total views: 47.4k
Views today: 0.47k
Verified
47.4k+ views
-Hint:- Convert all the given values to their S.I unit.
Now, evaluate the distance between the first and third minima.
Next, we can now calculate the width of slit by the expression –
$x = \dfrac{{\lambda D}}{d}$
where, $x$ is the distance between first and third minima,
$\lambda$ is the wavelength of light,
$D$ is the distance between screen and slit and
$d$ is the width of slit.

Complete Step by Step Solution:-
Let the distance between first and third minima be $x$ and first minima be ${x_1}$ and third minima be ${x_3}$. Therefore,
$x = {x_3} - {x_1} \\ x = 3mm \\ x = 3 \times {10^{ - 3}}m \\$
Let the distance between the screen and single slit be $D$. So, according to the question it is given that –
$D = 50cm \\ D = 0.5m \\$
Let the width of slit be $d$ and wavelength of light be $\lambda$.
According to the question, it is given that
$\lambda = 6000\dot A \\ \lambda = 6000 \times {10^{ - 10}}m \\$
The position of ${n^{th}}$ minima in the diffraction pattern is given by –
${x_n} = n\dfrac{{D\lambda }}{d}$
The above can also be rewritten by transposition as –
$d = n\dfrac{{D\lambda }}{x} \cdots (1)$
Putting the values of $x,n,D$ and $\lambda$ in their respective places in equation $(1)$
$d = (3 - 1)\dfrac{{0.5 \times 6000 \times {{10}^{ - 10}}}}{{3 \times {{10}^{ - 3}}}} \\ d = \dfrac{{2 \times 0.5 \times 6 \times {{10}^3} \times {{10}^{ - 10}}}}{{3 \times {{10}^{ - 3}}}} \\$
By further solving, we get
$d = \dfrac{{6 \times {{10}^{ - 4}}}}{3} \\ d = 2 \times {10^{ - 4}}m \\$
Now, converting metre to millimetre, we get
$d = 0.2mm$

Therefore, the correct answer is option (B).

Note:- Diffraction is a specialized case of scattering of light in which an object with regularly repeating features produces an orderly diffraction of light in a diffraction pattern. The diffraction plays an important role in limiting the resolving power of any optical instrument.
user
The text contains a physics problem that involves a mathematical calculation to determine the width of a slit based on given parameters. The problem is followed by a detailed step-by-step solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER