Question

The width of a plane incident wavefront is found
to be doubled in a denser medium. If it makes an
angle of 70° with the surface, calculate the
refractive index for the denser medium.​

Answer 1

Answer:

The angle of incidence, i = 70°

Since the width of refracted wave front is given double the width of the incident wave front i.e., from the figure attached below, we get CD = 2 AB.

∴ cos i / cos r = AB/CD

Or, cos 70° / cos r = AB / 2 AB

Or, 0.342 / cos r = ½  

Or, cos r = 0.684

Or, r = cos⁻¹ (0.684) = 46.842 ….. (i)

Now, we know that the formula for refractive index is given as,

µ = sin i / sin r  

or, µ = sin 70°/ sin (46.842) …… [from (i)]

or, µ = 0.939 / 0.729 = 1.28

Hence, the refractive index for the denser medium is 1.28.

Answer 2

Answer:

The angle of incidence which is given is i=70 degrees

As the girth of refracted wave front is given twofold the width of the incident wave front so we  

Let’s suppose,

∴ cos i / cos r = AB/CD

Or, cos 70° / cos r = AB / 2 AB

Or, 0.342 / cos r = ½  

Or, cos r = 0.684

Or, r = cos⁻¹ (0.684) = 46.842

Now, we see that the formula for refractive index is specified as,

µ = sin i / sin r  

or, µ = sin 70°/ sin (46.842

or, µ = 0.939 / 0.729 = 1.28

so, the refractive index for the denser medium is 1.28.