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Published on: 02/09/2022
QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Physics Subject - Retirement and Death of a Partner, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
Download Tamil Nadu 12th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
2.
Light transmitted by Nicol prism is, _____.
partially polarised
unpolarised
plane polarised
elliptically polarised
3.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,_____.
get shifted downwards
get shifted upwards
will remain the same
data insufficient to conclude
4.
A ray of light strikes a glass plate at an angle 60o. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is, _____.
\(\sqrt3\)
\(\frac{3}{2}\)
\(\sqrt{\frac{3}{2}}\)
2
5.
6.
When light is incident on a soap film of thickness 5 x 10–5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be, _____.
1.22
1.33
1.51
1.83
7.
Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are _____.
5I and I
5I and 3I
9I and I
9I and 3I
8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to, _____.
2D
\(\frac{D}{2}\)
\(\sqrt{2}\)D
\(\frac{D}{\sqrt2}\)
9.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is_____. [take wavelength of light, λ = 500 nm]
1 m
5 m
3 m
6 m
10.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is _____.
red
yellow
green
violet
1.
(d)
2.
(c)
plane polarised
3.
(b)
get shifted upwards
4.
n = tan ip = tan 60o = \(\sqrt{3}\)
5.
(b)
6.
2n t cos r = (2m + 1) \(\frac{\lambda}{2}\)
For maximum
m = 2 (For visible region), n - refractive index.
cos r = cos 0 = 1
t = 5 x 10-5 x 10-2 = 5 x 10-7 m
\(n=\frac{(2m+1)\frac{\lambda}{2}}{2t}=\frac{5\lambda}{2 \times 2 \times t}\)
\(=\frac{5 \times5320\times10^{-10}}{4 \times 5 \times 10^{-7}}\)
\(=\frac{5 \times5320\times10^{-10}}{20}=1330 \times 10^3\)
n = 1.330
7.
I = l1 + l2 + 2\(\sqrt{I_1I_2}\)cos θ
If cos θ = cos 0 = l, I is max
= I+ 4I + 2\(\sqrt{41^2}\) cos 0
= 5I + 4I = 91
If cos π = -1, I is min
Imin = I + 4I + 2\(\sqrt{41^2}\) cos π
= 5I + 4I(-1)
= 5I + 4I = I
(Imax, Imin)= (9I, I)
8.
d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)
9.
λ = 500 nm = 500 x 10-9 m
x = 3 mm = 3 x 10-3 m
a = 1 mm = 1 x 10-3 m
\(d=\frac{xa}{1.22 \lambda}\)
\(d=\frac{3 \times1\times10^{-6}}{1.22 \times500\times10^{-9}}\)
\(=\frac{3 \times1\times10^{-6}}{6.10 \times 10^{-7}}\)
\(d=\frac{30}{6.1}=5 m\)
10.
Refractive index for violet is more and wavelength for violet is very low comparing other colours. So, the letter which appears to be raised more is violet.
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