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Physics MCQs for NEET — Practice Questions with Answers

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The magnetic field in the plane electromagnetic wave is given by

$B_{z} = 2 \times 10^{- 7} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) t e s l a$

The expression for the electric field will be

$E_{z} = 30 \sqrt{2} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

$E_{z} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

$E_{y} = 30 \sqrt{2} \sin (0.5 \times 10^{11} x + 1.5 \times 10^{3} t) V / m$

$E_{y} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) V / m$

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Explanation

$\vec{B}$ along +z direction

EM wave travels along -x direction

Direction of $\vec{E} \times \vec{B}$ gives direction of wave

$∴ \vec{E}$ along y direction

Amplitude of electric field, $E_{0}$

$E_{0} = B_{0} c = 2 \times 10^{- 7} \times 3 \times 10^{8} = 60 V / m ∴ Electric field . E_{y} = 60 \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$

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For a transparent medium relative permeability and permittivity, $μ_{r} a n d ε_{r}$ are 1.0 and 1.44 respectively. The velocity of light in this medium would be:

2.5 $\times 10^{8}$ m/s

3 $\times 10^{8}$ m/s

2.08 $\times 10^{8}$ m/s

4.32 $\times 10^{8}$ m/s

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Explanation

$\begin{array}{l} μ = μ_{r} μ_{0} \\ ε = ε_{r} ε_{0} \\ v = \frac{1}{\sqrt{με}} \\ c = \frac{1}{\sqrt{μ_{0} ε_{0}}} \\ ∴ v = \frac{c}{\sqrt{μ_{r} ε_{r}}} = 2.5 \times 10^{8} m / s \end{array}$

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The minimum distance between a real object and screen such that a convex lens of focal length 20 cm can produce a real image on the screen is.

20 cm

40 cm

80 cm

60 cm

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Explanation

The minimum distance between the real object to the real image formed by the convex lens is 4f

Hence, D = 4f = 4 x 20 = 80 cm.

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A concave lens of focal length 25 cm produces image $\frac{1}{10} t h$ the size of the object. The distance of the object from the lens is

225 cm

250 cm

150 cm

175 cm

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Explanation

$m = \frac{v}{u} = \frac{1}{10} \Rightarrow v = \frac{u}{10} U \sin g l e n s f o r m u l a - \Rightarrow \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{10}{u} - \frac{1}{u} = \frac{1}{- 25} \Rightarrow u = - 225 c m$

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Assume that light of wavelength 7000 is coming from a star. The limit of resolution (in radian) of a telescope whose objective has a diameter of 244 cm, will be

$2.5 x 10^{- 7}$

$5.0 x 10^{- 7}$

$3.5 x 10^{- 7}$

$4.5 x 10^{- 7}$

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Explanation

$R . L . = \frac{1.22 λ}{a} = \frac{1.22 x 7 x 10^{- 7}}{2.44} R . L . = 3.5 x 10^{- 7}$

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A biconvex lens of glass $(μ = 1.5)$ has a radius of curvature 20cm. Which one of the following gives the correct position and nature of the image formed when an object is placed at 40cm from the lens?

Real, inverted and same size

real, inverted and diminished

Real, inverted and enlarged

Virtual, upright and enlarged

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Explanation

$(1) \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \frac{1}{f} = \frac{1}{2} (\frac{1}{20} + \frac{1}{20}) \frac{1}{f} = \frac{1}{2} x \frac{2}{20} f = 20 c m Object is placed a t 40 cm so image is also forme d at 40 cm from lens .$

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A converging beam of light is incident on a concave lens of focal length 30cm, in such a way that their point of intersection would lie at a height 1cm above the principal axis and at a distance of 24cm from the position of the lens if there was no lens. Find the nature of the image formed when the beam passes through the lens.

Real at a distance 60cm from the lens

Real at a distance 120cm from the lens

Virtual at a distance 60cm from the lens

Virtual at a distance 120 cm from the lens

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One face of a prism of apex angle 30 $°$ is silvered and a ray of light is incident on the first face at an angle i such that it retraces its entire path after reflection from the silvered face. If the refractive index of prism material is 2, then the value of i must be -

45 $°$

60 $°$

30 $°$

90 $°$

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Explanation

$A s t h e l i g h t r a y r e t r a c e s i t s p a t h a f t e r r e f l e c t i o n f r o m s i l v e r e d f a c e, r_{2} = 0 . r_{1} + r_{2} = A \Rightarrow r_{1} = A = 30^{0} μ = \frac{s i n i}{\sin r_{1}} \Rightarrow \sin i = μ \sin r_{1} = 2 \sin 30^{0} = 1 \Rightarrow i = 90^{0}$

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A ray of light incident on an equilateral prism at grazing incidence emerges from the prism at grazing emergence. Refractive index of the prism is

1.5

1.8

1.33

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Explanation

$G r a z i n g i n c i d e n t a n g l e = 90^{0} G r a z i n g e m e r g e n c e a n g l e = 90^{0} T h e n, r_{1} = r_{2} = \frac{A}{2} n = \frac{\sin i}{\sin r} = \frac{\sin 90^{0}}{\sin \frac{A}{2}} = \cos e c \frac{A}{2} = \cos e c 30^{0} = 2$

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The refractive index of the material of the prism for violet colour is 1.69 and that for red is 1.65. If the refractive index for mean colour is 1.66, the dispersive power of the material of the prism

0.66

0.06

0.65

0.69

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Explanation

$(2) ω = (\frac{μ_{v} - μ_{R}}{μ_{y} - 1}) = (\frac{1.69 - 1.65}{1.66 - 1}) = \frac{0.04}{0.66} = \frac{2}{33} = 0.06$

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