MCQ
There are certain material developed in laboratories which have a negative refractive index (Fig). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by:
- ✓
- B
- C
- D
✓
Answer
Correct option: A.
Solution:
The materials with negative refractive index responds to Snell’s law just opposite way. If incident ray from air (Medium 1) incident on those material, the ray refract or bend same side of the normal as in option (a).
Solution:
The materials with negative refractive index responds to Snell’s law just opposite way. If incident ray from air (Medium 1) incident on those material, the ray refract or bend same side of the normal as in option (a).
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The kinetic energy of an electron which is accelerated through a potential of $100$ volts is
→For the production of $X-$ rays, the target should be made of
→The drift velocity of electrons in silver wire with cross-sectional area $3.14 \times 10^{-6}\,m ^2$ carrying a current of $20\,A$ is. Given atomic weight of $Ag =108$ , density of silver $=10.5 \times 10^3\,kg / m ^3..........\times 10^{-4} m / sec$.
→What is not true of equipotential surface?
An alternating voltage of amplitude $40 \mathrm{~V}$ and frequency $4 \mathrm{kHz}$ is applied directly across the capacitor of $12 \mu \mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly:
→In $LCR$ circuit, the capacitance is changed from $C$ to $4C$. For the same resonant frequency, the inductance should be changed from $ L$ to
→The capacity of a parallel plate air condenser is $2 \mu F.$ If a dielectric of dielectric constant $4$ is introduced between the plates, its new capacity is:
→Referring to the figure below, the effective resistance of the network is
Which of the following statement is incorrect
Find the bindng energy per nucleon for $\frac{50}{120}\, Sn$. Mass of proton $m _{ P }=1.00783\, U ,$ mass of $m _{ n }=1.00867\, U$ and $mass$ of tin nucleus $m _{ Sn }=119.902199$ $U.$$......MeV$
→(take $1 U =931\, MeV )$