The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1st secondary maxima will be :
- 2 mm
- 2 cm
- 0.02 mm
- 0.2 mm
The Correct Option is A
Approach Solution - 1
The problem asks for the width of the first secondary maxima in a single-slit diffraction pattern. We are given the wavelength of light, the slit width, and the focal length of the convex lens used to focus the pattern.
Concept Used:
In a single-slit diffraction pattern, dark fringes (minima) are observed at angles \(\theta\) that satisfy the condition:
\[ a \sin(\theta) = m\lambda \]where \(a\) is the slit width, \(\lambda\) is the wavelength of light, and \(m = \pm 1, \pm 2, \pm 3, \ldots\) is the order of the minimum.
The secondary maxima are located approximately halfway between the minima.
The width of any secondary maxima is the distance between the two minima that bound it. The first secondary maxima is located between the first minimum (\(m=1\)) and the second minimum (\(m=2\)).
For small angles (which is a valid approximation in most diffraction experiments), \(\sin(\theta) \approx \tan(\theta) \approx \theta\). If the pattern is focused on a screen at a distance \(D\) from the slit (here, \(D\) is the focal length \(f\) of the lens), the position \(y\) of a fringe from the center is given by \(y = D \tan(\theta) \approx D\theta\).
So, the position of the \(m^{th}\) minimum is given by:
\[ y_m = \frac{m\lambda D}{a} \]The width of the first secondary maxima (\(W\)) is the difference in position between the 2nd and 1st minima:
\[ W = y_2 - y_1 \]Step-by-Step Solution:
Step 1: List the given values and convert them to SI units.
- Wavelength of light, \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\).
- Slit width, \(a = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}\).
- Focal length of the lens (distance to the screen), \(D = f = 100 \, \text{cm} = 1 \, \text{m}\).
Step 2: Find the position of the first minimum (\(y_1\)).
Using the formula for the position of the \(m^{th}\) minimum with \(m=1\):
\[ y_1 = \frac{1 \cdot \lambda D}{a} \] \[ y_1 = \frac{(400 \times 10^{-9} \, \text{m}) \times (1 \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ y_1 = \frac{400}{0.2} \times 10^{-6} \, \text{m} = 2000 \times 10^{-6} \, \text{m} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm} \]Step 3: Find the position of the second minimum (\(y_2\)).
Using the formula for the position of the \(m^{th}\) minimum with \(m=2\):
\[ y_2 = \frac{2 \cdot \lambda D}{a} \] \[ y_2 = \frac{2 \times (400 \times 10^{-9} \, \text{m}) \times (1 \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ y_2 = 2 \times y_1 = 2 \times (2 \times 10^{-3} \, \text{m}) = 4 \times 10^{-3} \, \text{m} = 4 \, \text{mm} \]Step 4: Calculate the width of the first secondary maxima.
The first secondary maxima lies between the first and second minima. Its width is the distance between these two minima.
\[ W = y_2 - y_1 \] \[ W = 4 \, \text{mm} - 2 \, \text{mm} = 2 \, \text{mm} \]Final Computation & Result:
The width of the first secondary maxima is the distance between the first and second dark fringes. Alternatively, the angular width of any secondary maxima is \( \Delta\theta = \frac{\lambda}{a} \). The linear width is \( W = D \Delta\theta = \frac{D\lambda}{a} \).
Using this direct formula:
\[ W = \frac{(1 \, \text{m}) \times (400 \times 10^{-9} \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} \] \[ W = \frac{400}{0.2} \times 10^{-6} \, \text{m} = 2000 \times 10^{-6} \, \text{m} = 2 \times 10^{-3} \, \text{m} \] \[ W = 2 \, \text{mm} \]The width of the 1st secondary maxima is 2 mm.
Approach Solution -2
The width of the first secondary maxima for single-slit diffraction is given by:
Width of 1st secondary maxima = \( \frac{\lambda D}{a} \)
where \( \lambda = 400 \times 10^{-9} \, \text{m} \), \( a = 0.2 \times 10^{-3} \, \text{m} \), and \( D = 100 \, \text{cm} = 1 \, \text{m} \). Substitute the values:
Width of 1st secondary maxima = \[ \frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 2 \, \text{mm} \]
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Optics Questions
- A parallel beam of monochromatic light of wavelength 5000 •Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by convex lens on screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of ________ (degree).
In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.
The focal length of the lens is_____cm.
- Two immiscible liquids of refractive indices \( \frac{8}{5} \) and \( \frac{3}{2} \) respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near-normal vision, the apparent depth of the coin is \( \frac{\alpha}{4} \) cm. The value of \( \alpha \) is ______.
- The refractive index of a prism with apex angle \( A \) is \( \cot \frac{A}{2} \). The angle of minimum deviation is:
- The following figure represents two biconvex lenses \( L_1 \) and \( L_2 \) having focal lengths 10 cm and 15 cm, respectively. The distance between \( L_1 \) and \( L_2 \) is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.