Test Index

CBSE Class 12 Physics 2023 Delhi Set 1 Paper

© examsnet.com
Question : 34 of 35
Marks: +1, -0
SECTION - E
A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, lens maker's formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u,vu, v and ff for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes.
(i) An object ABA B is kept in front of a composite convex lens, as shown in figure. Will the lens produce one image? If not, explain.
(ii) A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed? Explain.
(iii) A double convex lens is made of glass of refractive index 1.55 with both faces of the same radius of curvature. Find the radius of curvature required if focal length is 20  cm20\;\mathrm{cm}.
OR
(iii) Two convex lenses A and B of focal lengths 15  cm15\;\mathrm{cm} and 10  cm10\;\mathrm{cm} respectively are placed coaxially ' dd ' distance apart. A point object is kept at a distance of 30  cm30\;\mathrm{cm} in front of lens A. Find the value of ' dd ' so that the rays emerging from lens BB are parallel to its principal axis.
Solution:     👈: Video Solution
(i) No, the lens will not produce one image. It will produce two images. Refractive index upper and lower section of the lens being different there will be two foci and hence, two images will be formed.
(ii) If the screen is removed, the image will be still formed and it can be seen by looking through the lens through the specific cone of light.
(iii) Using lens maker's formula
  1f  =(μ−1)(  1r1−  1r2)\;\frac{1}{f}\;=(\mu-1)\left(\;\frac{1}{r_1}-\;\frac{1}{r_2}\right)
   Or,     120  =(1.55−1)(  2r)\;\text{ Or, }\;\;\frac{1}{20}\;=(1.55-1)\left(\;\frac{2}{r}\right)
∴  r  =22  cm\therefore\;r\;=22\;\mathrm{cm}
OR
The equivalent focal length of two lenses separated by a distance dd be ff
  1f=  1f1+  1f2−  df1f2\;\frac{1}{f}=\;\frac{1}{f_1}+\;\frac{1}{f_2}-\;\frac{d}{f_1 f_2}
Or,   1f=  115+  110−  d15×10\;\frac{1}{f}=\;\frac{1}{15}+\;\frac{1}{10}-\;\frac{d}{15 \times 10}
Or,   1f=  16−  d150\;\frac{1}{f}=\;\frac{1}{6}-\;\frac{d}{150}
When a point object is kept distance 30  cm30\;\mathrm{cm} from this lens combination, the rays emerge as parallel to the principal axis. So, focal length of the combination is 30  cm30\;\mathrm{cm}.
  ∴  130  =  16−  d150\;\therefore\;\frac{1}{30}\;=\;\frac{1}{6}-\;\frac{d}{150}
   Or,     d150  =  16−  130\;\text{ Or, }\;\;\frac{d}{150}\;=\;\frac{1}{6}-\;\frac{1}{30}
  ∴  d  =20  cm\;\therefore\;d\;=20\;\mathrm{cm}
© examsnet.com
Go to Question:
Prev Question
Next Question