English

A Ray of Light Falls on a Transparent Sphere with Centre C as Shown in the Figure. the Ray Emerges from the Sphere Parallel to the Line Ab.

Advertisements
Advertisements

Question

A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].

Advertisements

Solution

From Snell's law, we have: \[\frac{\sin\left( i \right)}{\sin\left( r \right)} = \mu\] At A, i = 60° (given)
Now, μ = √3

\[\Rightarrow \sin\left( r \right) = \frac{\sin\left( i \right)}{\mu}\]

\[ \Rightarrow \sin\left( r \right) = \frac{\sin\left( 60° \right)}{\sqrt{3}} = \frac{1}{2}\]

\[ \Rightarrow r = \sin^{- 1} \left( \frac{1}{2} \right)\]

\[ \therefore r = 30°\]

shaalaa.com
  Is there an error in this question or solution?
2013-2014 (March) Foreign Set 3

RELATED QUESTIONS

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?


Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.


 A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric. 


A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. 


A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.


A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.


A parallel-plate capacitor is filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.


A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:

1) the distance between the two plates is doubled?

2) a marble slab of dielectric constant 8 is introduced between the two plates?


Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.   


A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×