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Theory In the figure, i is increase...

Theory

In the figure, i is increased from `0^@ "to" 90^@`. But ray of light is travelling from denser to rarer medium. Therefore, TIR will take place when `igttheta_C`, where `theta_C=sin^-1(1/mu)=sin^-1(1/sqrt2)=45^@`
From `0^@-45^@`, refraction and reflection both will take place. After `45^@`, only reflection will take place.
Question Plot delta verses i graph between incident ray and refracted ray and, for `ile45^@` and with reflected ray for `ige45^@.`

Text Solution

Verified by Experts

For `ile45^@`
`delta=delta_("Refraction")=r-i` , (as `rgti`)
Applying Snell's law, we have
`mu=sinr/sini` or `sqrt2=sinr/sini`
`r=sin^-1(sqrt(2)sini)`
`:. r=sin^-1(sqrt2sini)`
Substituting this value of r in the equation, we have
`delta=sin^-1(sqrt(2)sini)-i`.......(i)
Now, variable are only two delta and i but this is not a known equation. So,
we can find some of the coordinates from where graph must pass.
At `i=0^@, delta=0^@`
At `i=45^@=theta_C,delta=45^@`
For `ige45^@`
`delta=delta_("Reflection")`
`:. delta=180^@-2i`
Now, delta verses i graph is a straight line.
At `i=45^@,delta=90^@`
At `i=90^@,delta=0^@`
Now, delta verses i graph is as shown in fig
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