CBSE Class 12 Physics 2022 Term 2 Delhi Set 1 Solved Paper

Question : 8

Total: 12

(a) (i) Define SI unit of power of a lens. 3
(ii) A plano- convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is 25cm.
(iii) Calculate the focal length of the lens.
(iv) If an object is placed 50cm in front of the lens, find the nature and position of the image formed.
OR
(b) A slit of width 0.6mm is illuminated by a beam of light consisting of two wavelengths 600nm and 480nm. The diffraction pattern is observed on a screen 1.0m from the slit. Find:3
(i) The distance of the second bright fringe from the central maximum pertaining to light of 600nm.
(ii) The least distance from the central maximum at which bright fringes due to both the wavelengths coincide.

Solution:

(i) SI unit of power of lens: Dioptre is the SI unit of power of a lens.
The power of the lens is the reciprocal of its focal length measured in metre(m).
(ii) (iii) Refractive index =mu=1.5
Radius of curvature of convex side =25cm
=

100

m
Radius of curvature of plane side =∞
Applying lens maker's formula,

=(μ−1)(

−

)
Or,

=(1.5−1)(

∞

100

)
Or,

=0.5×(

100

)
∴f=

m=50m
(iv) Focal Length =f=50cm
Object distance =u=−50cm
When the object distance = focal length, then the image will be formed at infinity.
Image will be highly magnified, real and inverted.
OR
(b) (i) Distance of 2nd bright fringe from central maximum =

2λD

2×600×10−9×1

0.6×10−3

(ii) If nth bright fringe due to 600nm coincides with (n+1) bright fringe due to 480nm, then

nλ1D

(n+1)λ2D

Or nλ1D=(n+1)λ2D
Or,

(n+1)

λ2

λ1

Or,

(n+1)

480

600

∴n=4
So, the least distance from central maximum
=

4×600×10−9×1

0.6×10−3

=40×10−4m

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