1.

A ray of light is incident on a surface of glass slab at an angle 45^(@). If the lateral shift produced per unit thickness is (1)/(sqrt(3))m, the angle of refraction prduced is :

Answer»

`ta^(-1)((SQRT(3))/(2))`
`TAN^(-1)(1-sqrt((2)/(3)))`
`SIN^(-1)(1-sqrt((2)/(3)))`
`tan^(-1)(sqrt((2)/(sqrt(3)-1)))`

Solution :(b) Here, angle of incidence `I = 45^(@)`

`("lateral SHIFT"(d))/("thickness of glass slab")=(1)/(sqrt(3))`
Angle of refraction, r = ?
`d=(tsindelta)/(cosr)=(TSIN(i-r))/(cosr)rArr(d)/(t)=(sin(i-r))/(cosr)`
`therefore(d)/(t)=(sinicosr-cosisinr)/(cosr)=sini-cositanr`
Substituting the values of the given quantities, we get
`(1)/(sqrt(3)) = sin 45^(@) - cos45^(@) tan r`
`(1)/(sqrt(3))=(1)/(sqrt(2))(1-tanr)rArr=tan^(-1)(1-sqrt((2)/(3)))`


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