Consider an equi-convex lens of glass left( m | vedclass

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(D) The focal length of an equi-convex lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{2}{R} \right)$.Since the focal

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$5 	imes 10^{-3}$

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(D) The focal length of an equi-convex lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{2}{R} ight)$.Since the focal length $f$ remains constant,$df = 0$. Differentiating the formula with respect to temperature,we get: $0 = d\mu \left( \frac{2}{R} ight) + (\mu - 1) \left( -\frac{2}{R^2} ight) dR$.Rearranging the terms,we have: $d\mu \left( \frac{2}{R} ight) = (\mu - 1) \left( \frac{2}{R^2} ight) dR$.Dividing both sides by $\frac{2}{R}$,we get: $d\mu = (\mu - 1) \frac{dR}{R}$.Since $\frac{dR}{R} = \alpha \Delta 	heta$,we substitute this into the equation: $d\mu = (\mu - 1) \alpha \Delta 	heta$.Given $\mu = 1.5$,$\alpha = 2.5 	imes 10^{-4} / ^\circ C$,and $\Delta 	heta = 40\, ^\circ C$.$d\mu = (1.5 - 1) 	imes (2.5 	imes 10^{-4}) 	imes 40$.$d\mu = 0.5 	imes 2.5 	imes 10^{-4} 	imes 40 = 0.5 	imes 100 	imes 10^{-4} = 50 	imes 10^{-4} = 5 	imes 10^{-3}$.

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