Question 283: Waves Optics

Question

In double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in interference pattern

Accepted Answer

When both slits have equal widths, they produce waves of equal intensity. If each slit provides an intensity $I$, the total resultant intensity at a point on the screen due to constructive interference (maxima) is:
$    I_{\max} = ( \sqrt{I} + \sqrt{I} )^2 = (2\sqrt{I})^2 = 4I.    $
For destructive interference (minima):
$    I_{\min} = ( \sqrt{I} - \sqrt{I} )^2 = 0.    $
So, with equal slit widths, we have:
Maxima intensity: $4I$
Minima intensity: $0$
Modified Scenario (One Slit Twice as Wide):
If one slit is made twice as wide as the other, the intensity from that slit will be greater. Intensity is proportional to the square of the amplitude and amplitude is proportional to the slit width.
Let's say the narrower slit still produces an intensity $I$. The wider slit, being twice as wide, will have twice the amplitude, hence:
$    	ext{Amplitude from narrower slit} = A    \implies 	ext{Intensity} = A^2 = I.    $
For the wider slit (double width):
$    	ext{Amplitude} = 2A \implies 	ext{Intensity} = (2A)^2 = 4A^2 = 4I.    $
Now we have two slit intensities: $I_1 = I$ (narrow slit) and $I_2 = 4I$ (wide slit).
Resultant Intensity in the Modified Case:
The resultant intensity at any point is:
$    I_{	ext{res}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi.    $
Substituting $I_1 = I$ and $I_2 = 4I$:
$    I_{	ext{res}} = I + 4I + 2\sqrt{I \cdot 4I}\cos\phi = 5I + 2 \cdot 2I \cos\phi = 5I + 4I\cos\phi.    $
Maximum and Minimum Intensities Now:
For maximum intensity ($\phi = 0^\circ$, $\cos\phi = 1$):
$      I_{\max} = 5I + 4I = 9I.      $
For minimum intensity ($\phi = 180^\circ$, $\cos\phi = -1$):
$      I_{\min} = 5I - 4I = I.      $
Previously, for equal slit widths:
Maxima = $4I$
Minima = $0$.
Now, after making one slit twice as wide:
Maxima = $9I$ (which is greater than the old maxima of $4I$)
Minima = $I$ (which is greater than the old minima of $0$).
Conclusion:
By making one slit wider, you unbalance the amplitudes. This causes:
Maxima intensity
to increase (from $4I$ to $9I$).
Minima intensity
to no longer be zero; it actually increases (from $0$ to $I$).
Answer:
The intensities of both the maxima and the minima increase.
Correct Option:
D) the intensities of both the maxima and the minima increase.

Answer

the intensity of the maxima decreases and the minima has zero intensity.

Answer

the intensity of maxima decreases and that of the minima increases.

Answer

the intensity of the maxima increases and the minima has zero intensity.

Answer

the intensities of both the maxima and the minima increase.

Waves Optics

In double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in interference pattern

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