If the effective focal length of Huygen-s eye piece is f-H- then the length of eye piece is -dfrac-3f-H-4-dfrac-4f-H-3-dfrac-2f-H-3-dfrac-3f-H-2-

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Toppr Admin · Accepted Answer

Two lenses of Huygen-apos-s eyepiece are of focal length f and 3f-Then- effective focal length of Huygens eyepiece -fH -3f2Length of eyepiece -2f-2-xD7-2fh3-4fH3

If the effective focal length of Huygen’s eye piece is $f_{H}$ , then the length of eye piece is :
$\frac{3 f_{H}}{4}$
$\frac{4 f_{H}}{3}$
$\frac{2 f_{H}}{3}$
$\frac{3 f_{H}}{2}$

Two lenses of Huygen's eyepiece are of focal length $f$ and $3 f$ .
Then, effective focal length of Huygens eyepiece $= f_{H}$ $= \frac{3 f}{2}$
Length of eyepiece $= 2 f = 2 \times \frac{2 f_{h}}{3} = \frac{4 f_{H}}{3}$

If the effective focal length of Huygen’s eye piece is fH, then the length of eye piece is :3fH44fH32fH33fH2

Two lenses of Huygen's eyepiece are of focal length f and 3f.Then, effective focal length of Huygens eyepiece =fH =3f2Length of eyepiece =2f=2×2fh3=4fH3

If the effective focal length of Huygen’s eye piece is $f_{H}$ , then the length of eye piece is :
$\frac{3 f_{H}}{4}$
$\frac{4 f_{H}}{3}$
$\frac{2 f_{H}}{3}$
$\frac{3 f_{H}}{2}$

Two lenses of Huygen's eyepiece are of focal length $f$ and $3 f$ .
Then, effective focal length of Huygens eyepiece $= f_{H}$ $= \frac{3 f}{2}$
Length of eyepiece $= 2 f = 2 \times \frac{2 f_{h}}{3} = \frac{4 f_{H}}{3}$