Answer
Verified
64.8k+ views
Hint: To solve this type of question firstly we have to find all the variables which are useful for finding the new fringe-width. So we will find the wavelength of light in water. And all other quantities are already given. Note that whenever the light source is placed in a different medium of the relative refractive index of $n$, there is a change in wavelength.
Formula used:
The formula for fringe width in YDSE is given by:
$\beta = \dfrac{{\lambda D}}{d}$
Complete step-by-step solution:
In Young's double-slit experiment a coherent light source illuminates a plate stabbed by two parallel slits, and the light passing through the slits is detected on a screen behind the plate. The wave nature of light causes the light waves to pass on through the two slits to interfere, creating bright and dark bands on the screen. The width of each fringe formed has a definite value across the entire band and it is known by the below formula.
The formula for fringe width in a double’s slit experiment is known by:
$\beta = \dfrac{{\lambda D}}{d}$
where,
$\lambda $ = wavelength of light
$D$ is the distance between the light source and the screen
$d$ is the distance between the slits.
Rearranging the equation, we can obtain
$\lambda = \dfrac{{\beta d}}{D}$ .......... $\left( 1 \right)$
When the whole experiment is conducted in water, there is a shift in the wavelength because whenever the light source is placed in a different medium of the refractive index of $n$ , there is a change in wavelength, given by:
$\lambda ' = \dfrac{\lambda }{n}$
Substitute the value of $\lambda $ in the equation $\left( 1 \right)$ , we have:
$\beta ' = \dfrac{{\lambda 'D}}{d}$
$\beta ' = \dfrac{{\beta d}}{{nD}} \times \dfrac{D}{d}$
On further solving the above equation we get,
$\beta ' = \dfrac{\beta }{n}$
Put the values that are already given in the question we get,
$\beta ' = \dfrac{\beta }{n}$
$ \Rightarrow \beta ' = \dfrac{{0.133}}{{1.33}} = 0.1cm$
Thus, the new fringe-width $\beta ' = 0.1cm$ .
Hence, the correct option is (B) $0.1\;cm$ .
Note: The wavelength of light in the denser medium will always be lesser than that of air since the refractive index of any denser medium like water, glass, etc. is greater than one. So, if you are doing the actual calculation with numbers, you should make sure that the wavelength in the denser medium should always be lesser, to ensure that you are moving in the right direction.
Formula used:
The formula for fringe width in YDSE is given by:
$\beta = \dfrac{{\lambda D}}{d}$
Complete step-by-step solution:
In Young's double-slit experiment a coherent light source illuminates a plate stabbed by two parallel slits, and the light passing through the slits is detected on a screen behind the plate. The wave nature of light causes the light waves to pass on through the two slits to interfere, creating bright and dark bands on the screen. The width of each fringe formed has a definite value across the entire band and it is known by the below formula.
The formula for fringe width in a double’s slit experiment is known by:
$\beta = \dfrac{{\lambda D}}{d}$
where,
$\lambda $ = wavelength of light
$D$ is the distance between the light source and the screen
$d$ is the distance between the slits.
Rearranging the equation, we can obtain
$\lambda = \dfrac{{\beta d}}{D}$ .......... $\left( 1 \right)$
When the whole experiment is conducted in water, there is a shift in the wavelength because whenever the light source is placed in a different medium of the refractive index of $n$ , there is a change in wavelength, given by:
$\lambda ' = \dfrac{\lambda }{n}$
Substitute the value of $\lambda $ in the equation $\left( 1 \right)$ , we have:
$\beta ' = \dfrac{{\lambda 'D}}{d}$
$\beta ' = \dfrac{{\beta d}}{{nD}} \times \dfrac{D}{d}$
On further solving the above equation we get,
$\beta ' = \dfrac{\beta }{n}$
Put the values that are already given in the question we get,
$\beta ' = \dfrac{\beta }{n}$
$ \Rightarrow \beta ' = \dfrac{{0.133}}{{1.33}} = 0.1cm$
Thus, the new fringe-width $\beta ' = 0.1cm$ .
Hence, the correct option is (B) $0.1\;cm$ .
Note: The wavelength of light in the denser medium will always be lesser than that of air since the refractive index of any denser medium like water, glass, etc. is greater than one. So, if you are doing the actual calculation with numbers, you should make sure that the wavelength in the denser medium should always be lesser, to ensure that you are moving in the right direction.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main