Light enters an isosceles right triangular prism at normal incidence through face AB and undergoes total internal reflection at face BC as shown below. The minimum value of the refractive index of the prism is close to:
A) $1.10$
B) $1.55$
C) $1.42$
D) $1.72$
Answer
Verified
119.7k+ views
Hint: By using Snell’s Law and knowing the terms of total internal reflection, we have to find the minimum value of the refractive index of the prism.
Total internal reflection: It is defined as the phenomenon which occurs when the ray of light travels from a more optically denser medium to a less optically denser medium. The angle of incidence must be greater than the critical angle. These are the conditions of total internal inflection.
Critical angle: The critical angle occurs when the angle of incidence where the angle of refraction is ${90^ \circ }$.
Formula Used:
We will be using the formula of Snell’s Law i.e., ${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Complete step by step solution:
Snell’s Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for a given pair of media and for the light of a given color. It gives the relation between the angle of incidence, the angle of refraction and refractive index of a given pair of mediums. It is given by,
${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Where,
${n_1}$,${n_2}$are the refractive index of medium $1$ and medium $2$
${\theta _1}$- Angle of incidence
${\theta _2}$- Angle of refraction.
Using Snell’s law,
${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Given: From the figure, ${n_1} = 1;{n_2} = ?;{\theta _1} = {90^ \circ };{\theta _2} = {45^ \circ }$
${n_2} = \dfrac{{sin{{90}^ \circ }}}{{\sin {{45}^ \circ }}}$
${n_2} = 1.414$
Hence the minimum value of the refractive index of the prism is close to $1.42$.
The answer is Option $(C)$, $1.42$.
Note: Snell’s law is derived from Fermat’s principle. It states that ‘’light travels in the shortest path that takes the least time’’. Snell’s Law gives the degree of refraction. The law was discovered by Dutch astronomer and Mathematician Willebrord Snell in $1621$.
Total internal reflection: It is defined as the phenomenon which occurs when the ray of light travels from a more optically denser medium to a less optically denser medium. The angle of incidence must be greater than the critical angle. These are the conditions of total internal inflection.
Critical angle: The critical angle occurs when the angle of incidence where the angle of refraction is ${90^ \circ }$.
Formula Used:
We will be using the formula of Snell’s Law i.e., ${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Complete step by step solution:
Snell’s Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for a given pair of media and for the light of a given color. It gives the relation between the angle of incidence, the angle of refraction and refractive index of a given pair of mediums. It is given by,
${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Where,
${n_1}$,${n_2}$are the refractive index of medium $1$ and medium $2$
${\theta _1}$- Angle of incidence
${\theta _2}$- Angle of refraction.
Using Snell’s law,
${n_1}sin{\theta _1} = {n_2}sin{\theta _2}$
Given: From the figure, ${n_1} = 1;{n_2} = ?;{\theta _1} = {90^ \circ };{\theta _2} = {45^ \circ }$
${n_2} = \dfrac{{sin{{90}^ \circ }}}{{\sin {{45}^ \circ }}}$
${n_2} = 1.414$
Hence the minimum value of the refractive index of the prism is close to $1.42$.
The answer is Option $(C)$, $1.42$.
Note: Snell’s law is derived from Fermat’s principle. It states that ‘’light travels in the shortest path that takes the least time’’. Snell’s Law gives the degree of refraction. The law was discovered by Dutch astronomer and Mathematician Willebrord Snell in $1621$.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation
Difference Between Circuit Switching and Packet Switching
Difference Between Mass and Weight
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
Sign up for JEE Main 2025 Live Classes - Vedantu
Trending doubts
Charging and Discharging of Capacitor
JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
Symbol of switch is ON position is class 12 physics JEE_Main
JEE Main 2025 Helpline Numbers for Aspiring Candidates
Electromagnetic Waves Chapter - Physics JEE Main
Other Pages
JEE Advanced 2024 Syllabus Weightage
A combination of five resistors is connected to a cell class 12 physics JEE_Main
JEE Main 2023 January 25 Shift 1 Question Paper with Answer Keys & Solutions
Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
JEE Mains 2025: Exam Dates, Updates, Eligibility and More
Christmas Day 2024 - Origin, History, and Why Do We Celebrate It