Question

3. A astronomical telescope has objective and
eyepiece of focal length 40 cm and 4 cm
respectively. To view an object 200 cm away from
the objective, the lenses must be separated by a
distance
[NEET-2016]
(1) 54.0 cm
(2) 37.3 cm
(3) 46.0 cm
(4) 50.0 cm​

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

➠ Focal length of objective lens = 40cm

➠ Focal length of eyepiece lens = 4cm

➠ Object distance for objective lens = -200cm

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

➝ At what distance should be lenses separated to view the object clearly.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

፨ Applying lens formula for objective lens.

$:\implies\tt\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$:\implies\tt\:\dfrac{1}{v}-\dfrac{1}{(-200)}=\dfrac{1}{40}$

$:\implies\tt\:\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{200}$

$:\implies\tt\:\dfrac{1}{v}=\dfrac{4}{200}$

$:\implies\boxed{\bf{\red{v=50\:cm}}}$

⇒ Image will be form at first focus of eyepiece lens.

⇒ So, for normal adjustment distance between objectives and eye piece lense will be

$\dashrightarrow\tt\:d=v+f_e$

$\dashrightarrow\tt\:d=50+4$

$\dashrightarrow\underline{\boxed{\bf\gray{d=54\:cm}}}$

NEETians (✊)