Answer
Verified
86.1k+ views
Hint: Use the formula of the maximum intensity and the minimum intensity to calculate it both of the light that enters the slit. Divide both to obtain the ratio of their intensities. In Young's experiment, the monochromatic light forms the different colors all over the screen.
Useful formula:
(1) The maximum intensity in the experiment is given by
${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$
Where ${I_{\max }}$ is the maximum intensity, ${I_1}$ is the intensity of the first slit and the ${I_2}$ is the intensity of the second slit.
(2) The minimum intensity is given by
${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$
Where ${I_{\min }}$ is the minimum intensity of the fringes.
Complete step by step solution:
It is given that
Reduction in the percentage of the intensity, if one slit is covered with glass, is $50\,\% $.
Let us assume that intensity of the slit one, ${I_1} = I$
So, the intensity of the slit two is ${I_2} = 0.5I$ (since the intensity reduces to $50\,\% $ )
By using the formula (1),
${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$
Substituting the values known,
${I_{\max }} = {\left( {\sqrt I + \sqrt {0.5I} } \right)^2}$
By simplifying,
${I_{\max }} = 2.9I$
Similarly using the formula (2),
${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$
${I_{\min }} = {\left( {\sqrt I - \sqrt {0.5I} } \right)^2}$
By simplification,
${I_{\min }} = 0.086I$
Ratio of the maximum and the minimum intensity is calculated as
$r = \dfrac{{{I_{\max }}}}{{{I_{\min }}}}$
$r = \dfrac{{2.9I}}{{0.086I}}$
By performing division in the above step,
$r = 33.8$
Thus the ratio of the maximum and the minimum intensity is $33.8$.
(b) Normally, the monochromatic light is used. But instead of this, if white light is used, the white fringe is formed at the center of the screen and the other colored fringes are positioned at various positions on the screen.
Note: The slits in the young’s double slit experiment are of the same distance from the center and of the same size. Hence the intensity of the light emitted by the slits $\left( I \right)$ are the same. But in this case , the second slit is covered with glass, so its intensity reduces $50\,\% $ . Hence the intensity of slit one is $I$ and slit two is $0.5I$ .
Useful formula:
(1) The maximum intensity in the experiment is given by
${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$
Where ${I_{\max }}$ is the maximum intensity, ${I_1}$ is the intensity of the first slit and the ${I_2}$ is the intensity of the second slit.
(2) The minimum intensity is given by
${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$
Where ${I_{\min }}$ is the minimum intensity of the fringes.
Complete step by step solution:
It is given that
Reduction in the percentage of the intensity, if one slit is covered with glass, is $50\,\% $.
Let us assume that intensity of the slit one, ${I_1} = I$
So, the intensity of the slit two is ${I_2} = 0.5I$ (since the intensity reduces to $50\,\% $ )
By using the formula (1),
${I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}$
Substituting the values known,
${I_{\max }} = {\left( {\sqrt I + \sqrt {0.5I} } \right)^2}$
By simplifying,
${I_{\max }} = 2.9I$
Similarly using the formula (2),
${I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}$
${I_{\min }} = {\left( {\sqrt I - \sqrt {0.5I} } \right)^2}$
By simplification,
${I_{\min }} = 0.086I$
Ratio of the maximum and the minimum intensity is calculated as
$r = \dfrac{{{I_{\max }}}}{{{I_{\min }}}}$
$r = \dfrac{{2.9I}}{{0.086I}}$
By performing division in the above step,
$r = 33.8$
Thus the ratio of the maximum and the minimum intensity is $33.8$.
(b) Normally, the monochromatic light is used. But instead of this, if white light is used, the white fringe is formed at the center of the screen and the other colored fringes are positioned at various positions on the screen.
Note: The slits in the young’s double slit experiment are of the same distance from the center and of the same size. Hence the intensity of the light emitted by the slits $\left( I \right)$ are the same. But in this case , the second slit is covered with glass, so its intensity reduces $50\,\% $ . Hence the intensity of slit one is $I$ and slit two is $0.5I$ .
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The thickness of the depletion layer is approximately class 11 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main