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Refraction of water waves - class-XI

Description: refraction of water waves
Number of Questions: 17
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Tags: oscillations and waves optics physics
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The higher the frequency of light?

  1. The longer (larger) its wavelength.

  2. The shorter (smaller) its wavelength.

  3. The greater its velocity in a vacuum.

  4. The redder it will be


Correct Option: B

A point source of light is placed at origin, in air. The equation of wave front of the wave at time $t$, emitted by source at $t=0$, is $(Take\ refractive\ index\ of\ air\ ms\ 1)$

  1. $x+y+z=et$

  2. $x^{2}+y^{2}+z^{2}=t^{2}$

  3. $xy+yz+zx=e^{2}t^{2}$

  4. $x^{2}+y^{2}+z^{2}=e^{2}t^{2}$


Correct Option: A

When the beam of light travels in a medium with lesser velocity than that of in vaccum then the value of wavelength and frequency will respectively

  1. Increase, decrease

  2. Increase, unchanged

  3. Decrease, unchanged

  4. Decrease, decrease


Correct Option: C
Explanation:

Energy $=\dfrac{4c}{\lambda }=4\nu $     which is constant.
Here, velocity is decreasing so, to make constant energy $\lambda$ should also decrease and frequency remains constant.

Consider a light wave that passes from air into a very thick clear glass block that has its opposite internal side mirrored (facing into the glass). The light ray passes into the glass block at an angle greater than $0^{\circ}$ and less than $90^{\circ}$, strikes the mirrored surface and reflects back through the glass into the air.
What happens while the light ray is in the glass block but before it strikes the mirrored surface?

  1. The frequency of the waves increases

  2. The frequency of the waves decreases

  3. The wavelength of the waves decreases

  4. The velocity of the waves decreases

  5. The period of the waves increases


Correct Option: C,D
Explanation:

When a light ray enters in a medium of higher refractive index from a lower one, the speed of the wave decreases as $v=\dfrac{c}{\mu}$

Since $\lambda=\dfrac{v}{\nu}$ and the frequency remains unchanged,
$\lambda=\dfrac{\lambda _0}{\mu}$
Hence the wavelength and the velocity of the waves decreases.

Assertion (A) : Velocity of light is more in rarer medium than in denser medium
Reason (R) : Light waves are longitudinal mechanical waves

  1. Statement A is correct R is wrong

  2. Statement A is correct R is right

  3. Statement A and R both are correct but R is not explaining A

  4. Statement A and R both are correct and R is explaining A


Correct Option: A
Explanation:

Velocity of light is more in rarer medium than in denser medium because speed of light in a medium is given by $\dfrac{c}{\mu }$ where, c is speed of light in vacuum and $\mu $ is refractive index of medium. Since denser medium has more $\mu $ than rarer medium, the speed of light is less in medium. Light waves are transverse electromagnetic waves.


(A) Light is a form of electromagnetic radiation and non-mechanical waves.
(B) Light waves do not require a material as a medium for propagation.

  1. Only A is true.

  2. Only B is true.

  3. Both A and B are true.

  4. Both A and B are false.


Correct Option: C
Explanation:

Both A and B are true,
Light is a form of electromagnetic radiation and a non-mechanical wave. It does not require a material as a medium for propagation. Hence, it can travel through space.

An observer is moving with half the speed of light towards stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer ?(speed of light = $3 \times10^{8} \ ms^{-1}$)

  1. 15.3 GHz

  2. 10.1 GHz

  3. 12.1 GHz

  4. 17.3 GHz


Correct Option: D
Explanation:
Doppler effect in light ( speed of observer is not very small compared to speed of light )
$f^{1} = \sqrt{\dfrac{1+V/C}{1-V/C}} f _{source} = \sqrt{\dfrac{1+1/2}{1-1/2}} (10\ GHz)$
$=17.3\ GHz$


In frounhofer diffraction by a single slit, a position where first order minimum is formed by the wave length $6000\mathring { A } $, first order maximum is formed due to unknown wavelength, the unknown wavelength is

  1. $5000\mathring { A } $

  2. $6000\mathring { A } $

  3. $4000\mathring { A } $

  4. $9000\mathring { A } $


Correct Option: C

Two monochromatic light source, A and B , emit the same number of photons wavelength of A is $ \lambda _A = 400 nm $, and that of $ \lambda _B =600 nm $. the power of radiated

  1. equal to that of source A

  2. less than that source A

  3. greater than that of source A

  4. can not be compared to that source A using the available data


Correct Option: C

When a ray of light enters from air in to water then its wavelength

  1. decreases

  2. increases

  3. remains unchanged

  4. becomes infinity


Correct Option: A
Explanation:

When light enters a denser medium from a rarer medium, its velocity decreases and hence the wavelength is reduced.

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 $E _n=-\cfrac{13.6}{n^2}eV)$
$E _2=-\cfrac{13.6}{2^2}eV\ \quad=-3.4eV\ \therefore\lambda=\cfrac{1240eVnm}{-3.4eV-(-13.6eV)}\ \quad=\cfrac{1240eVnm}{102eV}\ \quad=122nm$

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