
Light is incident from air on the surface of a glass slab, having a refractive index of\[1.5\]. In the air, the light beam makes an angle of \[{30^\circ }\] with the normal to the surface. What is the angle of the beam with the normal in the glass?
A.\[{\sin ^{ - 1}}\frac{2}{3}\]
B. \[{\sin ^{ - 1}}\frac{1}{3}\]
C. \[{90^ \circ }\]
D. None of the above
Answer
232.8k+ views
Hint: The phenomenon in which the ray of light travels from one medium to another is known as refraction of light. Due to change in medium, the speed of light changes accordingly. It either bends towards the normal or it bends away from the normal.
Formula Used:
Given that light is travelling from air to glass, therefore, the two mediums have different refractive index. Hence, Snell’s law will be used here. According to this law, the ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of mediums. Mathematically, it can be written as
\[\dfrac{{{\eta _2}}}{{{\eta _1}}} = \dfrac{{\sin i}}{{\sin r}} = \text{constant}\]
Where \[{\eta _1},{\eta _2}\] are the refractive indices of the two given mediums and $\sin i$, $\sin r$ is the angle of incidence and angle of refraction respectively.
Complete step by step solution:
Given that the angle of incidence is \[{\theta _1} = {30^ \circ }\].
It is also known that the refractive index of air is \[{\eta _1} = 1\].
Also it is given that the refractive index of glass is \[{\eta _2} = 1.5\].
Using the equation of Snell’s Law,
\[{\eta _1}\sin {\theta _1} = {\eta _2}\sin {\theta _2}\]
Substituting the given values and solving, the above equation can be written as
\[1 \times 0.5 = 1.5\sin {\theta _2} \\ \]
\[\Rightarrow \sin {\theta _2} = \dfrac{{1 \times 0.5}}{{1.5}} \\ \]
\[\Rightarrow \sin {\theta _2} = \dfrac{1}{3} \\ \]
\[\therefore {\theta _2} = {\sin ^{ - 1}}(\dfrac{1}{3})\]
Therefore, the angle of the beam with the normal in the glass is \[{\sin ^{ - 1}}(\dfrac{1}{3})\]
Hence, Option B is the correct answer.
Note:It is important to remember that when a ray of light travels from rarer to denser medium, the speed of light decreases and it bends away from the normal. On the other hand, when light ray travels from denser to rarer, its speed increases and it bends towards the normal.
Formula Used:
Given that light is travelling from air to glass, therefore, the two mediums have different refractive index. Hence, Snell’s law will be used here. According to this law, the ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of mediums. Mathematically, it can be written as
\[\dfrac{{{\eta _2}}}{{{\eta _1}}} = \dfrac{{\sin i}}{{\sin r}} = \text{constant}\]
Where \[{\eta _1},{\eta _2}\] are the refractive indices of the two given mediums and $\sin i$, $\sin r$ is the angle of incidence and angle of refraction respectively.
Complete step by step solution:
Given that the angle of incidence is \[{\theta _1} = {30^ \circ }\].
It is also known that the refractive index of air is \[{\eta _1} = 1\].
Also it is given that the refractive index of glass is \[{\eta _2} = 1.5\].
Using the equation of Snell’s Law,
\[{\eta _1}\sin {\theta _1} = {\eta _2}\sin {\theta _2}\]
Substituting the given values and solving, the above equation can be written as
\[1 \times 0.5 = 1.5\sin {\theta _2} \\ \]
\[\Rightarrow \sin {\theta _2} = \dfrac{{1 \times 0.5}}{{1.5}} \\ \]
\[\Rightarrow \sin {\theta _2} = \dfrac{1}{3} \\ \]
\[\therefore {\theta _2} = {\sin ^{ - 1}}(\dfrac{1}{3})\]
Therefore, the angle of the beam with the normal in the glass is \[{\sin ^{ - 1}}(\dfrac{1}{3})\]
Hence, Option B is the correct answer.
Note:It is important to remember that when a ray of light travels from rarer to denser medium, the speed of light decreases and it bends away from the normal. On the other hand, when light ray travels from denser to rarer, its speed increases and it bends towards the normal.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

