The optical properties of a medium are governed by the relative permittivity  (εr) and relative permeability (μr). The refractive index is defined as μ r ε r = n.For ordinary material εr > 0 and μr > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and μr < 0. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials n = – μ r ε r   . As light enters a medium of such refractive index the phases travel away from the direction of propagation.  (i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2nd quadrant, them the refracted beam is in the 3rd quadrant.  (ii) Prove that Snell’s law holds for such a medium.

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Explanation-All points with the same optical path length must have the same phase.

    So – μ r ε r A E   =BC-–  μ r ε r C D

    BC= μ r ε r (CD-AE)

    BC>0, si must be greater than AD

    But in other figure

    –  μ r ε r A E = B C     μ r ε r C D

    So BC= –  μ r ε r C D - A E

    But clearly here BE is less than zero

    To proving snells law we know that

    BC=ACsin θ and CD-AE=ACsin θ

    So n= sini/sinr

Similar Questions for you

A
alok kumar singh

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

R
Raj Pandey

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)

P
Payal Gupta

I=I0cos230°

=I0 (32)2=34I0 

V
Vishal Baghel

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

d y D = ( 2 n + 1 ) λ 2   [for Dark fringe at P]

n = 0, for first dark fringe

d y D = λ 2

λ = 2 d D y

= 2 d D × d 2 [ y = d 2 , G i v e n  first dark fringe is observed on the screen directly opposite to one of the slits]

λ = 2 × 0 . 6 × 0 . 6 8 0 0 × 2

λ = 4 5 0 m m

A
alok kumar singh

  Energy     Volume   = M L 2 T - 2 L 3 = M L - 1 T - 2

 The distance between two successive bright fringes is fringe width ( β ) .

β = λ D d = 589 × 10 - 9 × 1.5 0.15 × 10 - 3 = 5.9 m m

 

 

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