The focal lengths of the objective lens and t | vedclass

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(D) Given: Focal length of objective,$f_{o} = 30\, cm$.Focal length of eye lens,$f_{e} = 3.0\, cm$.The least distance of distinct vision,$D = 25\, cm$

Vedclass · Accepted Answer

$+8.8$

Vedclass · Answer

(D) Given: Focal length of objective,$f_{o} = 30\, cm$.Focal length of eye lens,$f_{e} = 3.0\, cm$.The least distance of distinct vision,$D = 25\, cm$.For a Galilean telescope,the magnifying power $M$ when the final image is formed at the least distance of distinct vision is given by:$M = \frac{f_{o}}{f_{e}} \left( 1 - \frac{f_{e}}{D} ight)$Substituting the values:$M = \frac{30}{3} \left( 1 - \frac{3}{25} ight)$$M = 10 	imes \left( \frac{25 - 3}{25} ight)$$M = 10 	imes \frac{22}{25}$$M = \frac{220}{25} = 8.8$Since the image is erect,the magnifying power is positive. Thus,$M = +8.8$.

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