Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Power of lenses of telescopes are 0.5D and 20D. If the final image is formed at least distance of distant vision(25cm), then what will be the length of the tube.

seo-qna
Last updated date: 14th Jun 2024
Total views: 392.4k
Views today: 11.92k
Answer
VerifiedVerified
392.4k+ views
Hint: Lens are the things which will allow the light to pass through them with convergence or divergence. Depending on whether the lens will converge or diverge the rays they are divided into convex and concave lenses. Their capacity of convergence or divergence is given by power of the lens.
Formula used:
$P = \dfrac{1}{f}$

Complete answer:
When an object is placed in front of the lens the rays from the object passes through the lens and suffers refraction and image is formed. The position where image forms and the nature of the image formed depends upon the type of lens we use.
Telescope is a device which is used to view distant objects very clearly.
Here in the telescope we use two lenses. Rays passing from the object will go through the convex lens first and suffer refraction. After that a refraction image will be formed which will be at some distance.
Now this image will act as an object for the second lens and image will be formed due to the second lens. This will be the final image seen by the observer. So the first convex lens is objective and the second convex lens is eye piece.
Power of the lens is given by the formula $P = \dfrac{1}{f}$
So the power of objective lens will be
$\eqalign{
  & {P_o} = \dfrac{1}{{{f_o}}} \cr
  & \Rightarrow 0.5 = \dfrac{1}{{{f_o}}} \cr
  & \therefore {f_o} = 2m = 200cm \cr} $
So the power of eyepiece lens will be
$\eqalign{
  & {P_e} = \dfrac{1}{{{f_e}}} \cr
  & \Rightarrow 20 = \dfrac{1}{{{f_e}}} \cr
  & \therefore {f_e} = 0.05m = 5cm \cr} $
Since we had got the focal lengths of the objective and eyepiece, we have formula to find the length of the telescope and that formula is
$L = {f_o} + \dfrac{{D{f_e}}}{{D + {f_e}}}$
Here ‘L’ is the length and ‘D’ is 25cm.
$\eqalign{
  & L = {f_o} + \dfrac{{D{f_e}}}{{D + {f_e}}} \cr
  & \Rightarrow L = 200 + \dfrac{{25(5)}}{{25 + 5}} \cr
  & \therefore L = 204.167cm \cr} $
So the length of the telescope is 204.167 centimeters.

Note:
If we observe clearly the powers of both the lens is positive. That means both are convex lenses. So the images formed by the convex lens are real or virtual. The focal length of the objective is greater than the focal length of the eye piece. If the image is formed at infinity then the sum of two focal lengths will be the length of the telescope.