The magnifying power of a telescope is 10 and length of telescope is 1.1 m for normal adjustment. The magnification when image is formed at least distance of distinct vision is -

Question

The magnifying power of a telescope is 10 and length of telescope is 1.1 m for normal adjustment. The magnification when image is formed at least distance of distinct vision is - (A) 14 (B) 6 (C) 16 (D) 18

Tardigrade Admin · Accepted Answer

10=(f0/fe) f0+fe ∴ f0=1 m f e=0.1 m When final image is at D; then mp = f 0((1/ D )+(1/ f e )) =100((1/25)+(1/10))=14

Q. The magnifying power of a telescope is 10 and length of telescope is $1.1 m$ for normal adjustment. The magnification when image is formed at least distance of distinct vision is -

14

6

16

18

$10 = \frac{f _{0}}{f _{e}} & f_{0} + f_{e}$
$∴ f_{0} = 1 m & f_{e} = 0.1 m$
When final image is at D; then
$m p = f_{0} (\frac{1}{D} + \frac{1}{f _{e}})$
$= 100 (\frac{1}{25} + \frac{1}{10}) = 14$

Q. The magnifying power of a telescope is 10 and length of telescope is 1.1m for normal adjustment. The magnification when image is formed at least distance of distinct vision is -

14

6

16

18

10=fe​f0​​&f0​+fe​ ∴f0​=1m&fe​=0.1m When final image is at D; then mp=f0​(D1​+fe​1​) =100(251​+101​)=14

Q. The magnifying power of a telescope is 10 and length of telescope is $1.1 m$ for normal adjustment. The magnification when image is formed at least distance of distinct vision is -

$10 = \frac{f _{0}}{f _{e}} & f_{0} + f_{e}$
$∴ f_{0} = 1 m & f_{e} = 0.1 m$
When final image is at D; then
$m p = f_{0} (\frac{1}{D} + \frac{1}{f _{e}})$
$= 100 (\frac{1}{25} + \frac{1}{10}) = 14$