JEE Physics Electromagnetic Waves MCQs Set 06

Answer: (b) Gamma rays

Explanation:
1. Gamma rays lie at the short wavelength end of the spectrum.

Teacher's Note:
1. Gamma rays have the shortest wavelength.

Answer: (b) \( 10^{14} \) Hz

Explanation:
1. The wavelength should be comparable to the size: \( \lambda\approx3\times10^{-6}\ \text{m} \).
2. \( f=\frac cl=\frac{3\times10^8}{3\times10^{-6}}=10^{14}\ \text{Hz} \)

Teacher's Note:
1. To see an object the wavelength must be of the order of its size.

Answer: (d) Radiowaves

Explanation:
1. Radio waves lie at the long-wavelength end of the spectrum.

Teacher's Note:
1. Radio waves have the longest wavelengths.

Answer: (a) d, b, a, c

Explanation:
1. Photon energy increases with frequency: radiowave < yellow < blue < X-ray.

Teacher's Note:
1. Energy per quantum is h times the frequency.

Answer: (d) For a wave propagating in the +x direction the electric field is \( \vec E=\frac1{\sqrt2}E_{yz}(\hat y-\hat z) \) and the magnetic field is \( \vec B=\frac1{\sqrt2}B_{yz}(\hat y+\hat z) \)

Explanation:
1. E, B and the propagation direction must be mutually perpendicular. In (a) and (b) the field is along or perpendicular wrongly; in (c) E and B are parallel.
2. In (d): \( (\hat y-\hat z)\times(\hat y+\hat z)=2\hat x \), perpendicular and along +x, so (d) is correct.

Teacher's Note:
1. Check that E cross B points along the propagation direction.

Answer: (b) \( (-2\hat i-3\hat j) \) and \( (3\hat i-2\hat j) \)

Explanation:
1. E and B must be perpendicular and \( \vec E\times\vec B \) must be along +z.
2. (a): perpendicular but cross product z-component is \( 1(-1)-2(2)=-5 \) (negative). (c): dot product 8, not perpendicular. (d): perpendicular, z-component \( 3(-3)-4(4)=-25 \) (negative).
3. (b): dot \( -6+6=0 \) and z-component \( (-2)(-2)-(-3)(3)=13 \) (positive), so (b).

Teacher's Note:
1. Use the dot product for perpendicularity and the z-component of the cross product for the direction.

Answer: (b) 2

Explanation:
1. Absorbed part gives pressure \( fS/c \); reflected part (1 - f) gives \( 2(1-f)S/c \).
2. \( P=\frac Sc[f+2(1-f)]=\frac Sc(2-f) \), so \( \lambda=2 \)

Teacher's Note:
1. Reflected energy transfers twice the momentum.

Answer: (c) 19

Explanation:
1. Heat needed \( =1\ \text{g}\times4.2\ \text{J/g}\,^\circ\text{C}\times1^\circ\text{C}=4.2\ \text{J} \); photon energy \( =\frac{1240}{575}\ \text{eV}=3.45\times10^{-19}\ \text{J} \).
2. \( N=\frac{4.2}{3.45\times10^{-19}}=1.2\times10^{19} \)

Teacher's Note:
1. Divide the heat required by the energy of one photon.

Answer: (b) 2

Explanation:
1. Displacement current equals the charging (conduction) current of the capacitor: \( I_D=C\frac{dV}{dt}=5\times10^{-6}\times2000=10^{-2}\ \text{A} \), so n = 2.

Teacher's Note:
1. Displacement current in the gap equals the rate of change of charge.

Answer: (c) 5

Explanation:
1. Energy incident \( =220\times\pi(0.3)^2\times300=1.87\times10^4\ \text{J} \).
2. Momentum \( =\frac Ec(0.6\times1+0.4\times2)=\frac{1.87\times10^4\times1.4}{3\times10^8}=8.71\times10^{-5} \), n = 5.

Teacher's Note:
1. Absorbed radiation gives E/c; reflected radiation gives 2E/c.

Answer: (b) 5

Explanation:
1. Intensity \( I=\frac12\epsilon_0cE_0^2=0.5\times8.85\times10^{-12}\times3\times10^8\times225=0.30\ \text{W/m}^2 \).
2. \( r=\sqrt{\frac{P}{4\pi I}}=\sqrt{\frac{100}{3.76}}\approx5.2\ \text{m} \)

Teacher's Note:
1. Equate the power divided by the sphere area to the intensity from the field amplitude.

Answer: (b) 714

Explanation:
1. \( I=\epsilon_0cE_{rms}^2\Rightarrow E_{rms}=\sqrt{\frac{1353}{8.85\times10^{-12}\times3\times10^8}}\approx714\ \text{V/m} \)

Teacher's Note:
1. Intensity equals epsilon0 c times the rms field squared.

Answer: (b) 13

Explanation:
1. Photons emitted per second \( =\frac{10}{3.37\times10^{-19}}=2.97\times10^{19} \).
2. Flux at 2 m \( =\frac{2.97\times10^{19}}{4\pi(200)^2\ \text{cm}^2}\approx5.9\times10^{13}\ \text{cm}^{-2}\text{s}^{-1} \), so P = 13.

Teacher's Note:
1. Divide the photon emission rate by the area of the sphere in cm squared.

Answer: (b) 2

Explanation:
1. \( k=\pi\times10^{-2}\Rightarrow\lambda=\frac{2\pi}{k}=200\ \text{m} \), so x = 2.

Teacher's Note:
1. Wavelength is 2 pi over k.

Answer: (c) 2

Explanation:
1. \( I=\frac12\epsilon_0cE_0^2=0.5\times8.85\times10^{-12}\times3\times10^8\times(38.8)^2\approx2.0\ \text{W/m}^2 \)

Teacher's Note:
1. Average intensity is one half epsilon0 c E0 squared.

Answer: (c) 4

Explanation:
1. \( \lambda=\frac cf=\frac{3\times10^8}{7.5\times10^6}=40\ \text{m} \) = 10 x 4, so x = 4.

Teacher's Note:
1. Wavelength is c divided by frequency.

Answer: (c) 4

Explanation:
1. \( B_0=\frac{E_0}{c}=\frac{120}{3\times10^8}=4\times10^{-7}\ \text{T} \)

Teacher's Note:
1. Divide the electric amplitude by c.