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Figure shows an irregular block of mater...

Figure shows an irregular block of material of refractive index `sqrt(2)`. A ray of light strikes the face AB as shown. After refraction, it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. Find the distance OE up to two places of decimal.

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From Snell's law ` mu_(1) sini =mu_(2)sinr` , we have
`sinr =(mu_(1))/(mu_(2))sini=(1)/(sqrt(2)) sin45^@=(1)/(2)rArr r=30^@`.
This mean that the ray becomes parallel to side AD inside the slab.
This implies that for the second face CD, `u=oo`
Given `R=0.4m`
Now, we have
`(mu_(3))/(v)-(mu_(2))/(u) =(mu_(3)-mu_(2))/(R)rArr (1.514)/(v)-(sqrt(2))/(oo)=(1.514-sqrt(2))/(R)`
`rArr v=(1.514xx0.4)/(1.514-1.414)=(1.514xx0.4)/(0.1)`
`=6.056m = 6.06m` (up to two decimal places).
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