When a light ray enters into glass medium from water at an angle of incidence 0º, what would be the angle of refraction?
0 ∘
Understanding Light Refraction at 0° Incidence
This question asks about the angle of refraction when a light ray travels from water into glass at a specific angle of incidence: 0°. Let's break down what happens when light passes from one medium to another.
What is Refraction?
Refraction is the bending of light as it passes from one transparent medium into another. This bending occurs because light travels at different speeds in different materials. The amount of bending depends on the properties of the two media and the angle at which the light hits the boundary between them.
The relationship between the angle of incidence (the angle between the incoming ray and the normal to the surface) and the angle of refraction (the angle between the refracted ray and the normal) is described by Snell's Law.
Applying Snell's Law to 0° Incidence
Snell's Law states:
\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)
Where:
- \(n_1\) is the refractive index of the first medium (water in this case).
- \(\theta_1\) is the angle of incidence (given as 0°).
- \(n_2\) is the refractive index of the second medium (glass in this case).
- \(\theta_2\) is the angle of refraction (what we want to find).
In this problem, the angle of incidence \(\theta_1\) is given as 0°. Let's plug this into Snell's Law:
\(n_1 \sin 0^\circ = n_2 \sin \theta_2\)
We know that the sine of 0° is 0 (\(\sin 0^\circ = 0\)). So the equation becomes:
\(n_1 \times 0 = n_2 \sin \theta_2\)
\(0 = n_2 \sin \theta_2\)
Since \(n_2\) (the refractive index of glass) is a non-zero value, for the product \(n_2 \sin \theta_2\) to be zero, \(\sin \theta_2\) must be zero.
\(\sin \theta_2 = 0\)
The angle whose sine is 0 is 0° (within the practical range of angles for refraction, which is typically 0° to 90°). Therefore, the angle of refraction \(\theta_2\) is 0°.
\(\theta_2 = 0^\circ\)
Physical Explanation
When a light ray strikes the boundary between two media perpendicularly, it is incident along the normal. In this specific case, the angle of incidence is 0°. When light is incident normally on a surface, it does not bend; it passes straight through into the second medium. This is true regardless of the refractive indices of the two media. The direction of the light ray does not change, meaning the angle of refraction is also 0°.
Analyzing the Options
Let's look at the given options based on our understanding:
- Option 1: 90°. This would mean the ray travels along the surface, which is not what happens at 0° incidence.
- Option 2: 45°. This implies bending, which doesn't occur at normal incidence.
- Option 3: 0°. This matches our derivation from Snell's Law and the physical principle of normal incidence.
- Option 4: The ray will not enter at all. This is incorrect; light generally enters a transparent medium unless conditions for total internal reflection are met (which is not the case here).
Based on the principles of refraction and Snell's Law, the angle of refraction is 0° when the angle of incidence is 0°.
| Parameter | Value |
|---|---|
| Medium 1 | Water |
| Medium 2 | Glass |
| Angle of Incidence (\(\theta_1\)) | 0° |
| Angle of Refraction (\(\theta_2\)) | ? |
| Snell's Law | \(n_1 \sin \theta_1 = n_2 \sin \theta_2\) |
Substituting \(\theta_1 = 0^\circ\):
\(n_1 \sin 0^\circ = n_2 \sin \theta_2\)
\(n_1 \times 0 = n_2 \sin \theta_2\)
\(0 = n_2 \sin \theta_2\)
Since \(n_2 \neq 0\), we must have \(\sin \theta_2 = 0\), which implies \(\theta_2 = 0^\circ\).
Conclusion on Angle of Refraction
When a light ray enters glass from water at an angle of incidence of 0°, it passes straight through without any deviation. Therefore, the angle of refraction is also 0°.
Revision Table: Light Refraction Basics
| Concept | Description |
|---|---|
| Refraction | Bending of light as it passes from one medium to another. |
| Angle of Incidence | Angle between incident ray and the normal at the point of incidence. |
| Angle of Refraction | Angle between refracted ray and the normal at the point of refraction. |
| Normal Incidence | When the angle of incidence is 0° (ray strikes perpendicular to the surface). |
| Snell's Law | Mathematical relation between angles of incidence/refraction and refractive indices: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\). |
| Refractive Index (n) | A measure of how much a medium slows down light; affects the amount of bending. |
Additional Information: Refraction and Optical Density
The bending of light during refraction is related to the change in optical density between the two media. Optical density is related to the refractive index. A medium with a higher refractive index is optically denser than a medium with a lower refractive index.
- When light travels from a rarer medium (lower \(n\)) to a denser medium (higher \(n\)), it bends towards the normal. The angle of refraction is less than the angle of incidence (\(\theta_2 < \theta_1\)).
- When light travels from a denser medium (higher \(n\)) to a rarer medium (lower \(n\)), it bends away from the normal. The angle of refraction is greater than the angle of incidence (\(\theta_2 > \theta_1\)).
However, these rules about bending towards or away from the normal apply only when the angle of incidence is not 0°. At 0° incidence, there is no bending, regardless of the change in optical density. The ray simply passes straight through the interface along the normal.
Similar Questions
Which of the following lenses will bend the light rays through the largest angle?
Which one of the following statements is not correct?
A person rings a metallic bell near a strong concrete wall. He hears the echo after 0.3 s. If the sound moves with a speed of 340 m/s, how far is the wall from him?
A sound wave has a frequency of 4 kHz and wavelength 30 cm. How long will it take to travel 2.4 km ?
If the wavelengths corresponding to ultraviolet, visible and infrared radiations are given as λ UV ,λ VIS and λ IR respectively, then which one of the following gives the correct relationship among these wavelengths?
The electromagnetic waves, which are used for satellite communication, are
The visible portion of the electromagnetic spectrum is __________.
A person can see near objects clearly through his eyes but cannot see distant objects clearly. The possible reason(s) for this defect could be:
1. Excessive curvature of the eye lens.2. The eye - ball has become too small.
3. The focal length of the eye lens is too large.
4. Elongation of the eye - ball.
Select the correct answer using the code given below:
Which one of the following statements is correct regarding the travel of a light beam from a rare to a dense medium?
- A person burned a firecracker in front of a cliff and heard its echo 5 s after it burst. The distance of the cliff from the person, if the speed of the sound is 340 m/s, is close to
Important Questions from Wave
Which of the following is correct?
I. Sound is a mechanical wave
II. Sound wave does not need any medium to propagate
The atmospheric green house effect is produced mainly by the absorption and re-emission of:
Which of the following are examples of electromagnetic waves?
a. Television waves
b. Ultraviolet rays
c. X-rays
d. Sun rays
When light passes from one medium into another medium, then the physical property which does not change is
The ray that comes back from a surface after the reflection is known as ______.
