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an apaque sphere of radius a is just immersed in a transparent liquid . Apoint sourse is placed on the vertical diameter of the sphere at a distance a/2 from the top ofthe sphere . one ray originating from the point source after refraction from the air liquidinterface forms tangent to the sohere. The angle of refraction for that particular ray is 30° .findthe refractive index of the liquid. .

Deepak Patra , 11 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

To find the refractive index of the liquid in which the opaque sphere is immersed, we can use Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media involved. In this case, we have air and the liquid. Let's break down the problem step by step.

Understanding the Setup

We have a sphere of radius a that is immersed in a transparent liquid. A point source is located on the vertical diameter of the sphere, specifically a/2 from the top of the sphere. A ray of light originating from this point source travels through the air, hits the air-liquid interface, and then refracts into the liquid. The ray is tangent to the sphere after refraction, and the angle of refraction is given as 30°.

Applying Snell's Law

Snell's Law states:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:

  • n₁ = refractive index of air (approximately 1)
  • θ₁ = angle of incidence in air
  • n₂ = refractive index of the liquid
  • θ₂ = angle of refraction in the liquid (given as 30°)

Determining the Angle of Incidence

Since the ray is tangent to the sphere after refraction, we can visualize that the angle of incidence θ₁ can be determined from the geometry of the situation. The tangent point creates a right triangle where:

  • The radius of the sphere is a.
  • The distance from the point source to the tangent point is a/2.

Using trigonometry, we can find that the angle of incidence θ₁ is equal to 90° - 30° = 60° because the angle of refraction is given as 30° and the tangent line is perpendicular to the radius at the point of tangency.

Calculating the Refractive Index

Now we can substitute the known values into Snell's Law:

1 * sin(60°) = n₂ * sin(30°)

We know:

  • sin(60°) = √3/2
  • sin(30°) = 1/2

Substituting these values gives:

√3/2 = n₂ * (1/2)

To isolate n₂, we multiply both sides by 2:

√3 = n₂

Final Result

The refractive index of the liquid is therefore √3, which is approximately 1.732. This value indicates how much the light slows down when it enters the liquid from air, and it helps us understand the optical properties of the liquid in relation to the sphere.

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