Question

pls answer as clear as possible​

Answer 1

$\boxed{\rm{\large{\red{Question\longrightarrow }}}}$

The given figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is:

$\boxed{\rm{\large{\green{SoLution\longrightarrow }}}}$

By using Snell's law,

Refractive index of medium B w.r.t. A is

$\boxed{ \sf{n= \dfrac{sin 60^0} {sin 45^0 } }}$

(as shown in the figure)

$\sf { n= \dfrac{\dfrac{\sqrt 3 }{2} }{\dfrac{1}{ \sqrt 2}} }$

here under root 2 is rational so we get,

$\bf {n=\dfrac{ \sqrt 3}{\sqrt 2} }$

$\sf{\pink{\underline{\bigstar \therefore Correct\ option \ is (a) \bigstar }}}$

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More info-

• snell's law - states the ratio of sine of angle of incidence to the sine of angle of refraction is always constant.
• $\bf{ Refractive \ index = \dfrac{sin_i}{sin_r}}$
• It is dimensionless unit. Have no SI unit.

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Answer 2

Answer:

By using Snell's law,

Refractive index of medium B w.r.t. A is

\boxed{ \sf{n= \dfrac{sin 60^0} {sin 45^0 } }}

n=

sin45

0

sin60

0

(as shown in the figure)

\sf { n= \dfrac{\dfrac{\sqrt 3 }{2} }{\dfrac{1}{ \sqrt 2}} }n=

2

1

23

here under root 2 is rational so we get,

\bf {n=\dfrac{ \sqrt 3}{\sqrt 2} }n=

23

★∴Correct option is(a)★

____________________________

More info-

snell's law - states the ratio of sine of angle of incidence to the sine of angle of refraction is always constant.

\bf{ Refractive \ index = \dfrac{sin_i}{sin_r}}Refractive index=

sin

r

sin

i

It is dimensionless unit. Have no SI unit.

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