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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 33 of 35
Marks: +1, -0
(a) (i) State Huygen's principle. With the help of a diagram, show how a plane wave is reflected from a surface. Hence, verify the law of reflection.
(ii) A concave mirror of focal length 12 cm12\,\text{cm} forms a three times magnified virtual image of an object. Find the distance of the object from the mirror.
OR
(b) (i) Draw a labelled ray diagram showing the image formation by a refracting telescope. Define its magnifying power. Write two limitations of a refracting telescope over a reflecting telescope.
(ii) The focal lengths of the objective and the eyepiece of a compound microscope are 1.0 cm1.0\,\text{cm} and 2.5 cm2.5\,\text{cm} respectively. Find the tube length of the microscope for obtaining a magnification of 300.
Solution:     👈: Video Solution
(a) (i) Huygens' principle:
Each point on primary wavefront act as a source of secondary wavelets sending out disturbances in all directions in the similar manner as the original source of light does.
Verification of laws of reflection:
Refer the answer of Q26 (OR) of CBSE 12 DELHI, Set-1.
   (ii)   \;\text{ (ii) }\;      Magnification     =−vu\;\; \text{ Magnification } \;\; = - \frac{v}{u}
     Or,     3=−vu\;\; \text{ Or, } \;\; 3 = - \frac{v}{u}
  ∴  v=−3u\; \therefore \; v = -3 u
Applying mirror formula,
    1v+1u=1f\;\; \frac{1}{v} + \frac{1}{u} = \frac{1}{f}
     Or,     −13u+1u=−112\;\; \text{ Or, } \;\; - \frac{1}{3u} + \frac{1}{u} = - \frac{1}{12}
     Or,     23u=−112\;\; \text{ Or, } \;\; \frac{2}{3} u = - \frac{1}{12}
  ∴  u=−8 cm\; \therefore \; u = -8 \,\text{cm}
(b) (i) Refer the answer of Q33 (OR) (i) of CBSE 12 DELHI, Set-1.
Limitation of refracting telescope:
● Refracting telescope suffer from chromatic aberration.
● Large size Refracting telescopes are too heavy and too costly due to large size of lenses
(ii) Given,
f0=1 cmf_0 = 1 \,\text{cm}
FE=2.5 cmF_E = 2.5 \,\text{cm}
m=300m = 300
   Since,     m=L×Df0×fE\; \text{ Since, } \;\; m = \frac{L \times D}{f_0 \times f_E}
   or,     300=L×251×2.5\; \text{ or, } \;\; 300 = \frac{L \times 25}{1 \times 2.5}
  ∴  L=30 cm\; \therefore \; L = 30 \,\text{cm}
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