![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
The central fringe of interference pattern produced by light of wavelength $6000$ A Is found to shift to the position of 4th bright fringe, after a glass plate of $\mu = 1.5$ is introduced. The thickness of the glass plate is:
A) $4.8\mu m$
B) $8.23\mu m$
C) $14.98\mu m$
D) $3.78\mu m$
Answer
418.2k+ views
Hint: Shift at ${n^{th}}$ bright fringe is $(\mu - 1)t = n\lambda $ (where, $\mu $ is the refractive index of the glass plate, $t$ is the thickness of the glass plate, $\lambda $ is the wavelength of the light).
Complete step by step answer:
When two light waves of the same frequency and amplitude superpose in a certain region of a medium, the intensity of the resultant light wave increases at certain points and decreases at some other points in that region. This phenomenon is known as the interference of light and these alternate dark and bright lines are called interference fringes.
According to the formula, shift at ${n^{th}}$ bright fringe is $(\mu - 1)t = n\lambda $ (where, $\mu $ is the refractive index of the glass plate, $t$ is the thickness of the glass plate, $\lambda $ is the wavelength of the light.)
Putting all the given values in the equation, we get,
$(1.5 - 1)t = 4(6000 \times {10^{ - 10}})$
$\implies t = \dfrac{{4(6000 \times {{10}^{ - 10}})}}{{0.5}} = 4.8 \times {10^{ - 10}}m = 4.8\mu m$
So, the thickness of the glass plate is $4.8\mu m$.
$\therefore$ The correct option is A.
Additional Information: In an interference pattern, there is no loss or destruction of light energy in the dark fringes area. The energy just gets shifted from the region of the dark band to the region of the bright band. Total energy remains the same. It can be shown that the average intensity of a set of simultaneous consecutive dark and bright fringes is the same as the intensity of the usual illumination in the same region. Hence, we can say that the interference fringe does not contradict the law of conservation of energy.
Note: It is to be noted that, when light travels through a medium, the equivalent optical path of the actual path traveled by light is the product of the refractive index of the medium and the actual path traveled. Also, though a displacement occurs in fringe pattern due to insertion of a glass plate, fringe width remains unaltered.
Complete step by step answer:
When two light waves of the same frequency and amplitude superpose in a certain region of a medium, the intensity of the resultant light wave increases at certain points and decreases at some other points in that region. This phenomenon is known as the interference of light and these alternate dark and bright lines are called interference fringes.
According to the formula, shift at ${n^{th}}$ bright fringe is $(\mu - 1)t = n\lambda $ (where, $\mu $ is the refractive index of the glass plate, $t$ is the thickness of the glass plate, $\lambda $ is the wavelength of the light.)
Putting all the given values in the equation, we get,
$(1.5 - 1)t = 4(6000 \times {10^{ - 10}})$
$\implies t = \dfrac{{4(6000 \times {{10}^{ - 10}})}}{{0.5}} = 4.8 \times {10^{ - 10}}m = 4.8\mu m$
So, the thickness of the glass plate is $4.8\mu m$.
$\therefore$ The correct option is A.
Additional Information: In an interference pattern, there is no loss or destruction of light energy in the dark fringes area. The energy just gets shifted from the region of the dark band to the region of the bright band. Total energy remains the same. It can be shown that the average intensity of a set of simultaneous consecutive dark and bright fringes is the same as the intensity of the usual illumination in the same region. Hence, we can say that the interference fringe does not contradict the law of conservation of energy.
Note: It is to be noted that, when light travels through a medium, the equivalent optical path of the actual path traveled by light is the product of the refractive index of the medium and the actual path traveled. Also, though a displacement occurs in fringe pattern due to insertion of a glass plate, fringe width remains unaltered.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the difference between Conduction and conv class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Mark the correct statements about the friction between class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the acceleration of the wedge towards the right class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A standing wave is formed by the superposition of two class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Derive an expression for work done by the gas in an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Class 11 JEE Main Physics Mock Test 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
![arrow-right](/cdn/images/seo-templates/arrow-right.png)