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The figure shows a ray incident a angle ...

The figure shows a ray incident a angle `i=pi//3` of the plot drawn shown the variartion of `|r-i|`versus`(mu_(1))/(mu_(2))`=k,`(r="angle of refraction")`

the value of`k_(1)`is

A

The value of `k_(1) "is" (2)/(sqrt(3))`

B

The value of `theta_(1) "is" pi//6`

C

The value of `theta_(2) "is" pi//3`

D

The value of `k_(0) "is" 1`

Text Solution

Verified by Experts

The correct Answer is:
B,C,D


`delta = 0 at K_(0) implies K_(0) = (mu_(1))/(mu_(2)) = 1`

`theta_(2) = ? , K = (mu_(1))/(mu_(2)), mu_(1) : "denser,K : "increases"`
r =0 , `" " delta= (pi)/(3)-0 implies theta_(2) = (pi)/(3)`
`theta_(1) = ? " " K = (mu_(1))/(mu_(2)) downarrow`
`mu_(2)` : denser, `mu_(1)`:rarer: K : decreases
`mu_(2)"sin"i = mu_(1)"sin"(pi)/(2)`
`"sin"(pi)/(3) = (mu_(1))/(mu_(2)) = (sqrt(3))/(2) = K_(1)`
At `k_(1) , r = (pi)/(2), so, theta_(1) = (pi)/(2)- (pi)/(3) = (pi)/(6)`
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