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Glass has refractive index μ with respect to air and the critical angle for a ray of light going from glass to air is an if a ray of light is incident from the air on the glass with the angle of incidence, a corresponding angle of refraction is
- a)sin-1(μ)
- b)sin-1(1/μ2)
- c)sin-1(1/√μ)
- d)sin-1 (1/μ)
Correct answer is option 'B'. Can you explain this answer?
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Glass has refractive index μ with respect to air and the critical ang...
Refractive index of glass with respect to air = μ Critical angle = θ
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∴ μ = (1/sinθ) …(i)
When a ray of light incident from the air on the glass with the angle of incidence θ. If r be the angle of refraction, then by Snell’s law,
μ = sinθ/sinr
⇒ 
[fromEq. (i)]
[fromEq. (i)]⇒ sinr = 1/μ2r = sin-1(1/μ2)
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Glass has refractive index μ with respect to air and the critical ang...
Explanation:
When a ray of light is incident from air to glass, it bends towards the normal because the speed of light is less in glass than in air. The angle of refraction is given by Snell's law,
μsin(i) = sin(r)
where μ is the refractive index of the glass with respect to air, i is the angle of incidence and r is the angle of refraction.
When the angle of incidence i is greater than or equal to the critical angle, the angle of refraction becomes 90 degrees and the ray of light is totally reflected back into the glass. The critical angle is given by the equation,
sin(a) = 1/μ
where a is the critical angle.
Now, let us consider the case when the ray of light is incident from air to glass with an angle of incidence i. The corresponding angle of refraction r is given by Snell's law as,
μsin(i) = sin(r)
We need to find the value of r. To do this, we can use the relationship between the sine of an angle and its complementary angle. We have,
sin(90 - r) = cos(r)
Substituting sin(r) from the Snell's law equation, we get
sin(90 - r) = μsin(i)
Using the trigonometric identity sin(90 - x) = cos(x), we get
cos(r) = μsin(i)
Solving for r, we get
r = sin^-1(μsin(i))
Since sin(i) = 1/μsin(a) (using Snell's law), we can substitute sin(i) in the above equation to get
r = sin^-1(sin(a))
Using the identity sin(a) = 1/μ, we get
r = sin^-1(1/μ)
Therefore, the corresponding angle of refraction is sin^-1(1/μ), which is option (B).
When a ray of light is incident from air to glass, it bends towards the normal because the speed of light is less in glass than in air. The angle of refraction is given by Snell's law,
μsin(i) = sin(r)
where μ is the refractive index of the glass with respect to air, i is the angle of incidence and r is the angle of refraction.
When the angle of incidence i is greater than or equal to the critical angle, the angle of refraction becomes 90 degrees and the ray of light is totally reflected back into the glass. The critical angle is given by the equation,
sin(a) = 1/μ
where a is the critical angle.
Now, let us consider the case when the ray of light is incident from air to glass with an angle of incidence i. The corresponding angle of refraction r is given by Snell's law as,
μsin(i) = sin(r)
We need to find the value of r. To do this, we can use the relationship between the sine of an angle and its complementary angle. We have,
sin(90 - r) = cos(r)
Substituting sin(r) from the Snell's law equation, we get
sin(90 - r) = μsin(i)
Using the trigonometric identity sin(90 - x) = cos(x), we get
cos(r) = μsin(i)
Solving for r, we get
r = sin^-1(μsin(i))
Since sin(i) = 1/μsin(a) (using Snell's law), we can substitute sin(i) in the above equation to get
r = sin^-1(sin(a))
Using the identity sin(a) = 1/μ, we get
r = sin^-1(1/μ)
Therefore, the corresponding angle of refraction is sin^-1(1/μ), which is option (B).
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Glass has refractive index μ with respect to air and the critical angle for a ray of light going from glass to air is an if a ray of light is incident from the air on the glass with the angle of incidence, a corresponding angle of refraction isa)sin-1(μ)b)sin-1(1/μ2)c)sin-1(1/√μ)d)sin-1 (1/μ)Correct answer is option 'B'. Can you explain this answer?
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