JEE Main 2019 (Online) 12th January Evening Slot | Electromagnetic Waves Question 115 | Physics

A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :

6 V/m

10 V/m

30 V/m

24 V/m

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

-1

A 27 mW laser beam has a cross-sectional area of 10 mm². The magnitude of the maximum electric field in this electromagnetic wave is given by :
[Given permittivity of space $$ \in $$₀ = 9 $$ \times $$ 10^–12 SI units, Speed of light c = 3 $$ \times $$ 10⁸ m/s]

2 kV/m

1 kV/m

1.4 kV/m

0.7 kV/m

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

An electromagnetic wave of intensity 50 Wm^–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

$$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$$

$$\left( {\sqrt n ,\sqrt n } \right)$$

$$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$$

$$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$$

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