Answer
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.
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.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
Other Pages
An electric dipole is placed in an electric field generated class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main