Question

show that the refractive index of plastic is 1.43.Given that sin 40°=0.64 and sin44°=0.69​

Answer 1

Explanation:

In Δ QDR

δ=i−r+(i

′

+r

′

)

=(i+i

′

)−(r+r

′

) .....(1)

In Quadrilateral AQER ∠A+∠E=180

o

.....(2)

In Δ QER r+r

′

+∠E=180

o

............(3)

r+r

′

=∠A [from equations (2) and (3)]

Putting value of r+r

′

in equation (1)

δ=1+i

′

−∠A ........(4)

In the position of minimum deviation condition.

i=i

′

,r=r

′

,δ=δ

m

,

So, r+r

′

=∠A

2r=∠A

or r=∠A/2 .............(5)

Equations (4) and (5) becomes

δ

m

=2i∠A

or i=

2

δA+δ

m

........(6)

Putting value of i and r from (5),(6), in Snell's law

n=

sinr

sini

n=

sinA/2

sin

2

(A+δ

m

)

A=Angle of deviation

δ

m

=Minimum angle.

solution

Answer 2

Given:

Given that, the material is plastic.

Also,  $\[\mathf{sin}\text{ }\mathf{40}{}^\circ =\mathf{0}.\mathf{64}\]$ and $\[\mathf{sin}\text{ }\mathf{44}{}^\circ =\mathf{0}.\mathf{69}\]$

To find:

We need to find the refractive index of plastic.

Solution:

The ratio of a medium is defined as how the light travels through that medium.

It’s a dimensionless measure. It defines what proportion a light – weight ray will be bent when it enters from one medium to the opposite.

Snell’s law clarifies the relation between the angle of incidence and angle of refraction. So, there are two formulas for calculating the index of refraction of a medium. Allows us to imagine that a ray of sunshine is traveling from a medium a to a different medium b.

Then conferring from Snell’s law,

$\[n=\frac{\sin i}{\sin r}\]$

Formula of index of refraction is $n=\frac{sin i}{sin r}$,

$i=40{}^\circ ,r=44{}^\circ$

Substituting the values within the above equation,

$\[n=\frac{\sin 40{}^\circ }{\sin 44{}^\circ }\]$

$n=\frac{0.64}{0.69}$

When solving this equation, we get the worth of ratio index of plastic,

$\[n=0.925\approx1$