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A glass slab is placed on a page of a book kept horizontally. What should be the value of the minimum refractive index of the glass slab so that the printed letters of the page will not be visible from any vertical side of the slab? |
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Answer» Solution :It is assumed that there is a thin layer of air between the page of the book and the glass SLAB. So ray coming from any portion of the book is incident on the glass at an ANGLE of almost `90^(@)` [Fig. 2.62]. If the angle of REFRACTION is `phi` and refractive index of glass then, `sinphi = (1)/(mu)` If `theta_(c)` is the critical angle, then `sintheta_(c) = (1)/(mu)`. So here, `phi = theta_(c)` This REFRACTED ray is incident on any vertical side of the slab at an angle `theta(say). So" " theta + phi = 90^(@)` If `theta` is greater than `phi`, total internal reflection of light takes place and the printed letters of te page will not be visible from any vertical side. `mu` will be minimum when `theta = phi`. `therefore "" 2phi = 90^(@) or,phi = 45^(@)` `therefore "" mu_(min) = (1)/(sin phi) = (1)/(sin 45^(@)) = sqrt(2) = 1.414` 
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