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The figure shows a transparent slab of l...

The figure shows a transparent slab of length m placed in air whose refractive index in x direction varies as `mu = 1 + x^(2)(0 lt x lt 1)`. The optical path length of ray R will be

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`(a) 4/3 mm , (b) 1/12 m`
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The lens governing the behavior of the rays namely rectilinear propagation laws of reflection and refraction can be summarised in one fundamental law known as Fermat's principle. According to this principle a ray of light travels from one point to another such that the time taken is at a stationary value (maximum or minimum). if c is the velocity of light in a vacuum the velocity in a medium of refractive index mu is (c)/(mu) hence time taken to travel a distance l is (mul)/(c) if the light passes through a number of media, the total time taken is ((1)/(c))summul or (1)/(c)intmudl if refractive index varies continuously. Now summul is the total path, so that fermat's principle states that the path of a ray is such that the optical path in at a stationary value. this principle is obviously in agreement with the fact that the ray are straight lines i a homogenous isotropic medium. it is found that it also agrees with the classical laws of reflection and refraction. Q. If refractive index of a slab varies as mu=1+x^(2) where x is measured from one end then optical path length of a slab of thickness 1 m is

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