1.

Consider a transparent hemispher (n=2) in front of which a small object isplaced in air (n=1) as shown in figure. Q. Consider a ray starting from O which strikes the spherical surface at grazing incident (i=90^(@)). Takin x=R, what will be the angle (from normal) at which the ray may emerge from the plane surface.

Answer»

`90^(@)`
`0^(@)`
`30^(@)`
`60^(@)`

Solution : i. `(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`
Taking refraction first at curved surface,
`(2)/(v_(1))+(1)/(X)=(1)/(R)rArr v_(1)=(2Rx)/(x-R)`
For plane surface,
`v^(')=v_(1)RrArrv^(')=(xR+R^(2))/(x-R)`
`rArr (1)/(v)-(2(x-R))/(R(x+R))=0`
` (1)/(v)-(2(x-R))/(R(x+R))`
For virtual IMAGE,
`(1)/(v)lt0rArr (2(x-R))/(R(x+R))lt0`
`x lt R`
II. For `x=2R`
`V_(1)=(4R^(2))/(R)= 4RrArru=-2R`
`m_(1)=(mu_(1))/(mu_(2)),(v)/(u)=(1)/(2), (4R)/((-2R))=-1`
`m_(2)=1rArr m_(1)m_(2)=-1`
Image is real, inverted, and of same size.
(iii) Hence, correct answer is `90^(@)`


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