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Physics · 2022 · 3 marks
CBSE 2022 · Region 4 · Set 1 · Q8
(a) (i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope. (ii) A telescope consists of an…
(i)
Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope.
(ii)
A telescope consists of an objective of focal length $\displaystyle 150$ cm and an eyepiece of focal length $\displaystyle 6.0$ cm. If the final image is formed at infinity, then calculate :
(I)
the length of the tube in this adjustment, and
(II)
the magnification produced.
(i)
Draw a labelled ray diagram showing the formation of the image at least distance of distinct vision by a compound microscope.
(ii)
A small object is placed at a distance of $\displaystyle 3 \cdot 0$ cm from a magnifier of focal length $\displaystyle 4.0$ cm. Find :
(I)
the position of the image formed, and
(II)
the linear magnification produced.
Marking-scheme solution
(i)
(ii)
Given $\displaystyle f_o = 150$ cm, $\displaystyle f_e = 6$ cm
(I)
Length of the tube $\displaystyle L = f_o + f_e$
$\displaystyle = 150 + 6$
$\displaystyle L = 156$ cm
(II)
$\displaystyle m = \dfrac{f_o}{f_e}$
$\displaystyle \dfrac{150}{6} = 25$
(i)
Ray diagram of image formation by a compound microscope
(ii)
Given $\displaystyle u = -3$ cm, $\displaystyle f = 4$ cm
(I)
Using $\displaystyle \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
$\displaystyle \dfrac{1}{v} = \dfrac{1}{u} + \dfrac{1}{f} = \dfrac{1}{-3\cdot0} + \dfrac{1}{4\cdot0}$
$\displaystyle \dfrac{1}{v} = \dfrac{-4+3}{12} = -\dfrac{1}{12}$ $\displaystyle v = -12$ cm
(II)
Linear magnification $\displaystyle m = \dfrac{v}{u}$
$\displaystyle m = \dfrac{-12}{-3} = 4$
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