Tag: huygen's wave theory

Questions Related to huygen's wave theory

The wavelength of a monochromatic light in vacuum is $\lambda$. If travels from vacuum to a medium of absolute refractive index $\mu$. The ratio of wavelength of the incident and refracted wave is

  1. $\mu^2 : 1$

  2. 1 : 1

  3. $\mu$ : 1

  4. 1:$\mu$


Correct Option: C
Explanation:

${ \left{ \mu \lambda  \right}  } _{ vacuum }={ \left{ \mu \lambda  \right}  } _{ medium }\ \Rightarrow \dfrac { { \lambda  } _{ i } }{ { \lambda  } _{ r } } =\dfrac { \mu  }{ 1 } \ \therefore \ Ratio\ is\ \mu :1$

The frequency of light of wave length 5000 $\mathring {A}$ is

  1. $1.5 \times 10^5 Hz$

  2. $6 \times 10 Hz$

  3. $6 \times 10^{14} Hz$

  4. $7.5 \times 10^{15} Hz$


Correct Option: C
Explanation:

Answer is B.

The relation between the velocity of light, frequency and wavelength is given as v = f$\lambda $.
The velocity of light v = $3 \times 10^{8}m/s$.
So, f = $v/\lambda $ = $3 \times 10^{8}m/s / 5000\mathring{A}$ = $6 \times 10^{14} Hz$.
Hence, the frequency of the light is $6 \times 10^{14} Hz$.

When a ray of light is refracted, the wavelength of the refracted light changes. Identify which of the following explain this phenomenon?
I. Some of the energy of the incident ray is carried away by the reflected ray
II. The boundary surface absorbs some of the energy of the incident ray
III. The incident and refracted rays do not travel with the same velocity

  1. I only

  2. II only

  3. III only

  4. I and II only

  5. I, II, and III


Correct Option: C
Explanation:
Wavelength of the light in a medium $\lambda _m = \dfrac{v _m}{\nu}$ where $v _m$ is the velocity of light in that medium
The velocity with which the incident and the refracted ray travels is not same but the frequency of both the rays is same. Hence the wavelength of the refracted light changes w.r.t the incident light.
Hence option C is correct.

Calculate the wavelength of light when it passing through diamond whose refractive index is 2.4 given that wavelength of light in air is 500 nm.

  1. 208 nm

  2. 357 nm

  3. 500 nm

  4. 700 nm

  5. 1,200 nm


Correct Option: A
Explanation:

Given :   $n = 2.4$              $\lambda _a = 500$ nm

$\therefore$ Wavelength of light in diamond        $\lambda _d = \dfrac{\lambda _a}{n}  =\dfrac{500}{2.4} =208$  nm

An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines?

  1. Frequency of emitted radiation

  2. Wave length of emitted radiation

  3. Energy of emitted radiations

  4. Velocity of light


Correct Option: A

Ray optics holds good when characteristic dimensions are

  1. Of the same order as the wavelength of light

  2. Much smaller than the wavelength of light

  3. Of the order of one millimetre

  4. Much larger than the wavelength of light


Correct Option: D
Explanation:

Ray optics holds good when characteristic dimensions are much larger than the wavelength of light.

What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum. (take hc = 1240 eV nm)

  1. 102 nm

  2. 150 nm

  3. 82 nm

  4. 122 nm


Correct Option: D
Explanation:

Find out $\lambda$ of light

$hc=1240eVmm$
Lymen series of the hydrogen spectrum
Solution
For any series the transistion that produces the least energetic photon is the transition between the home base level that defines the series and the level immediately above it. For the hymen series, the home-base level is at $n=1$
So the transition that produces the least energetic photon is the transition from the $n=2$ level to then $n=1$ level.
$\therefore$ The wavelength for the least energetic photon is
$\lambda=\cfrac{hc}{E _2-E _1}$
Here $hc=1240eVnm$
$E _1=-\cfrac{13.6}{1^2}eV\ \quad=-13.6eV$
(as $E _n=-\cfrac{13.6}{n^2}eV)$
$E _2=-\cfrac{13.6}{2^2}eV\ \quad=-3.4eV\ \therefore\lambda=\cfrac{1240eVnm}{-3.4eV-(-13.6eV)}\ \quad=\cfrac{1240eVnm}{102eV}\ \quad=122nm$