Question 254: Waves Optics

Question

In a Young's double slit experiment wavelength of light used is $6000 \mathop {\rm{A}}\limits^{\rm{o}}$. The first order maxima and tenth order maxima fall at 14.50 mm and 16.75 mm from the particular reference point in the interference pattern respectively. If the wavelength is changed to $5500 \mathop {\rm{A}}\limits^{\rm{o}}$ then the position of zero order and tenth order maxima are respectively
[The other arrangements remaining same]

Accepted Answer

Step 1: Recall the YoungÃ¢ÂÂs double-slit fringe position formula
In YoungÃ¢ÂÂs double slit experiment, the position of the $ m^{	ext{th}} $ order maximum is given by:
$ y_m = m \frac{\lambda D}{d} $
where
$ m $ = order of the fringe,
$ \lambda $ = wavelength,
$ D $ = distance between slits and screen,
$ d $ = slit separation.
Thus,
the fringe position is proportional to $ m \lambda $
for fixed $ D/d $.
Step 2: Use the given information
We are told that:
First order maximum ($ m=1 $) at $ \lambda = 6000 \, 	ext{ÃÂ} $ lies at $ y_1 = 14.50 \, 	ext{mm}$.
Tenth order maximum ($ m=10 $) at same wavelength lies at $ y_{10} = 16.75 \, 	ext{mm}$.
So, fringe positions are not measured from the central fringe (zero-order maxima), but from some
reference point
.
Step 3: Find the difference
For wavelength $ 6000 \, 	ext{ÃÂ} $:
$ y_{10} - y_1 = 16.75 - 14.50 = 2.25 \, 	ext{mm}. $
This distance corresponds to
9 fringes
(from $ m=1 $ to $ m=10 $).
So, fringe width is:
$ \Delta y = \frac{2.25}{9} = 0.25 \, 	ext{mm}. $
Step 4: Locate zero-order maximum with original wavelength
Since for $ \lambda = 6000 \, 	ext{ÃÂ} $, each fringe spacing is 0.25 mm:
First order maximum is at 14.50 mm.
Therefore,
zero-order maximum
is 1 fringe earlier:
$ y_0 = 14.50 - 0.25 = 14.25 \, 	ext{mm}. $
So the zero-order maximum lies at 14.25 mm.
Step 5: Check tenth order
From calculated zero position:
$ y_{10} = y_0 + 10 \Delta y = 14.25 + (10 	imes 0.25) = 14.25 + 2.50 = 16.75 \, 	ext{mm}. $
Ã¢ÂÂ Matches the given.
So reference point is consistent.
Step 6: For new wavelength $ \lambda = 5500 \, 	ext{ÃÂ} $
Fringe width (direct proportionality to wavelength):
$ \Delta y' = \Delta y 	imes \frac{5500}{6000} = 0.25 	imes \frac{5500}{6000}. $
$ \Delta y' = 0.2292 \, 	ext{mm (approx)}. $
Step 7: Zero-order position
The
zero order maximum position
does not depend on wavelength (itÃ¢ÂÂs central bright fringe), so it remains at:
$ y_0' = 14.25 \, 	ext{mm}. $
Step 8: Tenth-order position with new wavelength
$ y_{10}' = y_0' + 10 	imes \Delta y' = 14.25 + 10 	imes 0.2292. $
$ y_{10}' = 14.25 + 2.292 \approx 16.55 \, 	ext{mm}. $
Ã°ÂÂÂ Final Answer:
Zero-order and tenth-order maxima positions are:
$ \boxed{14.25 \, 	ext{mm}, \, 16.55 \, 	ext{mm}} $
That corresponds to
Option A
. Ã¢ÂÂ

Answer

$14.25 \mathrm{~mm}, 16.55 \mathrm{~mm}$

Answer

$12.25 \mathrm{~mm}, 14.55 \mathrm{~mm}$

Answer

$10.25 \mathrm{~mm}, 12.55 \mathrm{~mm}$

Answer

$16.25 \mathrm{~mm}, 18.55 \mathrm{~mm}$

$$\mathop {\rm{A}}\limits^{\rm{o}} $$

Waves Optics

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