A parallel beam of light is incident on a thi | vedclass

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(A) The deviation $\delta$ produced by a thin prism is given by $\delta = (\mu - 1)A$.Given $\mu = 1.5$ and $A = \frac{4}{\pi}$ degrees.First,convert

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$\left( 60 \ cm, \frac{2}{3} \ cm \right)$

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(A) The deviation $\delta$ produced by a thin prism is given by $\delta = (\mu - 1)A$.Given $\mu = 1.5$ and $A = \frac{4}{\pi}$ degrees.First,convert the angle $A$ to radians: $A = \left( \frac{4}{\pi} ight) 	imes \left( \frac{\pi}{180} ight) = \frac{1}{45} \, rad$.Now,calculate the deviation: $\delta = (1.5 - 1) 	imes \frac{1}{45} = 0.5 	imes \frac{1}{45} = \frac{1}{90} \, rad$.The light rays,after passing through the prism,are deviated by an angle $\delta$ and then pass through a convex lens of focal length $f = 60 \ cm$.The converging point (focus) of the beam will be at a distance $f$ from the lens along the deviated path.The vertical displacement $y$ from the principal axis is $y = f 	imes \delta = 60 	imes \frac{1}{90} = \frac{2}{3} \, cm$.Thus,the coordinates of the converging point are $\left( 60 \ cm, \frac{2}{3} \ cm ight)$.

$A$ parallel beam of light is incident on a thin prism with a prism angle of $\frac{4}{\pi}$ degrees. The refractive index of the prism is $1.5$. The focal length of the lens is $60 \ cm$. The coordinates of the converging point of the beam are:

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