
A ray of light is incident at the glass-water interface at an angle i. It merges finally parallel to the surface of the water. Then, the value of ${\mu _g}$ would be (where ${\mu _g}$ is the refractive index of glass with respect to water)

A. $\left( {\dfrac{4}{3}} \right)\sin i$
B. $\dfrac{1}{{(\sin i)}}$
C. $\dfrac{4}{3}$
D. $1$
Answer
161.4k+ views
Hint: Here in this question, given that a ray of light is incident through glass medium to the water medium at an angle i and merges parallel to the surface of water. Here we have to find the value of ${\mu _g}$ . As we know that we can easily find the solution using Snell’s law.
Formula used:
Snell’s law
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}}$
Complete answer:
As we know that by applying the Snell’s law, we can easily find the solution of the question,
By applying the Snell’s law at the position of Glass-water surface,
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}}$
Here, $w$ stands for water and $g$ stands for glass,
As by putting the value we get the value as,
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}} = \dfrac{{{}_g\mu }}{{{}_w\mu }}....(i)$
As we take the above equation as equation(i)
Now, Applying the Snell’s law at the position of water-air surface,
${}_w^a\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin r}}$
Here, $a$ stands for air and $w$ stands for water,
By putting the whole value in above equation, we get,
${}_w^a\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin r}} = \dfrac{{{}_w\mu }}{{{}_a\mu }}....(ii)$
Now, by multiplying equation(i) with equation(ii) we get,
$\dfrac{{{}_g\mu }}{{{}_a\mu }} = \dfrac{1}{{\sin i}}$
By taking the whole conclusion we get,
${}_g\mu = \dfrac{1}{{\sin i}}$
Therefore, the correct answer for the value of ${}_g\mu $ is $\dfrac{1}{{\sin i}}$ .
Hence, the correct option is (B).
Note:
Snell's law often applies to isotropic or specular material only (such as glass). The ordinary or o-ray, which obeys Snell's law, and the other extraordinary or e-ray, which may not be co-planar with the incident ray, can be separated by birefringence in anisotropic media, such as certain crystals. So, check the material before using the formula.
Formula used:
Snell’s law
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}}$
Complete answer:
As we know that by applying the Snell’s law, we can easily find the solution of the question,
By applying the Snell’s law at the position of Glass-water surface,
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}}$
Here, $w$ stands for water and $g$ stands for glass,
As by putting the value we get the value as,
${}_g^w\mu = \dfrac{{\sin r}}{{\sin i}} = \dfrac{{{}_g\mu }}{{{}_w\mu }}....(i)$
As we take the above equation as equation(i)
Now, Applying the Snell’s law at the position of water-air surface,
${}_w^a\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin r}}$
Here, $a$ stands for air and $w$ stands for water,
By putting the whole value in above equation, we get,
${}_w^a\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin r}} = \dfrac{{{}_w\mu }}{{{}_a\mu }}....(ii)$
Now, by multiplying equation(i) with equation(ii) we get,
$\dfrac{{{}_g\mu }}{{{}_a\mu }} = \dfrac{1}{{\sin i}}$
By taking the whole conclusion we get,
${}_g\mu = \dfrac{1}{{\sin i}}$
Therefore, the correct answer for the value of ${}_g\mu $ is $\dfrac{1}{{\sin i}}$ .
Hence, the correct option is (B).
Note:
Snell's law often applies to isotropic or specular material only (such as glass). The ordinary or o-ray, which obeys Snell's law, and the other extraordinary or e-ray, which may not be co-planar with the incident ray, can be separated by birefringence in anisotropic media, such as certain crystals. So, check the material before using the formula.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Wheatstone Bridge for JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
