A rectangular slab ABCD, of refractive index `n_1` , is immersed in water of refractive index `n_2(n_1 lt n_2)`. A ray of light is incident at the surface AB of the slab as shown in Fig. Find the maximum value of angle of incidence `alpha_(max)`, such that the ray comes out only from the other surface CD.

Options

A. `sin^-1[n_1/n_2cos(sin^-1.n_2/n_1)]`
B. `sin^-1[n_1cos(sin^-1.(1)/(n_2))]`
C. `sin^-1(n_1/n_2)`
D. `sin^-1(n_2/n_1)`

Answer

A

Explanation

`n_1sinC=n_2sin90^@impliessinC=n_2//n_1`
`C=sin^-1(n_2//n_1)`
Refraction at AB: `n_2sinalpha=n_1sinr`
`sinr=n_2/n_1sinalpha` …(i)
`r+i=90^@impliesr=90^@-i`
For T.I.R. on AD:
`igeC`
`sinigesinCimpliessinigen_2//n_1`
`igesin^-1(n_2//n_1)`
`cosigecos[sin^-1(n_2//n_1)]`
`sinrlecos[sin^-1(n_2//n_1)]`
`n_2/n_1sinalphalecos[sin^-1(n_2//n_1)]`
`sinalphalen_1/n_2cos[sin^-1(n_2//n_1)]`
`alphalesin^-1[n_1/n_2cos{sin^-1(n_2//n_1)}]`
`alpha_(max)=sin^-1[n_1/n_2cos{sin^-1(n_2//n_1)}]`