A light ray parallel to the optical axis of a glass sphere of radius R enters the sphere at height H above the axis (H<R).
The ray is refracted from air with ¥è = 34¨¬ into the sphere of unknown refractive index.
If the refracted ray emerges at the point P where the sphere intercepts the optical axis, what is the refractive index of the sphere?
How do you define the optical axis of a sphere ? What is the meaning of your
è and ¨¬ characters?
¥è = 34¨¬
I think it turned into different characters after copying and pasting it onto the website. I think the ¥è represented "theta" (which usually symbolizes an angle) and ¨¬ actually stood for degrees for an angle.
To find the refractive index of the glass sphere, we can use the concept of refraction. Refraction occurs when light passes through a medium with a different refractive index, causing the light ray to change direction.
In this scenario, we have a light ray that is parallel to the optical axis of the glass sphere. We know the refractive index of air (¥è = 34¨¬) and the height (H) at which the light ray enters the sphere.
To determine the refractive index of the glass sphere, we can use Snell's Law, which relates the angles of incidence and refraction and the refractive indices of the two mediums:
n1 * sin(θ1) = n2 * sin(θ2)
where:
n1 is the refractive index of the initial medium (air),
θ1 is the angle of incidence (which is zero for a ray parallel to the optical axis),
n2 is the refractive index of the second medium (glass),
and θ2 is the angle of refraction.
Since the ray is parallel to the optical axis, the angle of incidence is zero degrees. Therefore, sin(θ1) = 0.
The equation simplifies to:
0 = n2 * sin(θ2)
Since sine cannot be zero (except for at θ2 = 0), we can conclude that the refractive index of the glass sphere is also zero.
Therefore, the refractive index of the glass sphere cannot be determined with the given information.