Given a slab with index n=1.33 and incident light striking the top horizontal face at angle i as shown in figure. The maximum value of i for which total internal reflection occurs is

Options

A. `sin^(-1)(sqrt(0.77))`
B. `cos^(-1)(sqrt(0.77))`
C. `sin^(-1)(0.77)`
D. `sin^(-1)(0.38)`

Answer

A

Explanation

(a) TIR to take place ,
`90^(2)-rgttheta_(C)`
`rArrrgt90^(@)-theta_(C)`
From the relation,
`mu=(sini)/(sinr)`
`sini=musinr`
Since, for TIR to take place,`rlt90^(@)-theta_(C)`
`therefore i=sin^(-1){musinr}`
`ilesin^(-1)[{nsin(90^(@)-theta_(C))}]`
or `ilesin^(-1){mucostheta_(C)}`
or `ilesin^(-1){musqrt(1-1/(mu^(2)))} (becausesintheta_(C)=1/(mu_(C)))`
or `ilesin^(-1){1.33sqrt(1-1/((1.33)^(2)))}`
or `ilesin^(-1){sqrt(0.77)}`