Tag: wave optics

Questions Related to wave optics

In Fresnel's biprism expt., a mica sheet of refractive index 1.5 and thickness 6 $\times$ 10$^{-6}$m is placed in the path of one of interfering beams as a result of which the central fringe gets shifted through 5 fringe widths. The wavelength of light used is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 6000 $\overset{o}{A}$

  2. 8000 $\overset{o}{A}$

  3. 4000 $\overset{o}{A}$

  4. 2000 $\overset{o}{A}$

Show answer
Correct Option: A
Explanation:

Where n is equivalent number of fringe by which the centre fringe is shifted due to mica sheet
$\displaystyle\lambda = \frac{(\mu - 1)t}{n} = \frac{(1.5 - 1) 6 \times 10^{-6}}{5} = 6 \times 10^{-7} m = 6000 \overset{o}{A}$

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $\dfrac {a^{2}}{\lambda}$

  2. $\dfrac {\lambda}{a^{2}}$

  3. $\dfrac {\lambda}{a}$

  4. $\dfrac {\lambda^{2}}{a}$

  5. $a^{2}\lambda$

Show answer
Correct Option: A
Explanation:

An aperture of size a is illuminated by a parallel beam of light of wavelength $\lambda$. The distance at which ray optics has a good approximation is $\dfrac {a^{2}}{\lambda}$. This is the Fresnel distance.

Diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

The statement is true.

That is diffraction gratings provide much brighter interference pattern since more light passes through them compared with double slits.
One difference between the interference of many slits (diffraction grating) and double-slit (Young's Experiment) is that a diffraction grating makes a number of principle maxima along with lower intensity maxima in between.  The principal maxima occur on both sides of the central maximum for which a formula similar to Young's formula holds true.

In biprism experiment, the distance of 20th bright band from the centre of the interference pattern is 8 mm. The distance of the 30th bright band is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. 4 mm

  2. 8 mm

  3. 12 mm

  4. 16 mm

Show answer
Correct Option: C
Explanation:

Given

$20\cfrac { \lambda D }{ d } =8\ =\cfrac { \lambda D }{ d } =\cfrac { 8 }{ 20 } \ \therefore  { 30 }^{ th }$ bright,
$=30\times \cfrac { \lambda D }{ d } \ =30\times \cfrac { 8 }{ 20 } \quad =12mm$

In a biprism experiment, the fringe width obtained on the screen is $6\ mm$ from the slits which are $1.5\ m$ away from each other. Keeping the setting unchanged if the eye-piece is moved $20\ cm$ towards the biprism, find the change in fringe width.

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $0.90\ mm$

  2. $0.86\ mm$

  3. $0.80\ mm$

  4. $0.53\ mm$

Show answer
Correct Option: A

In a Fresnel's bi-prism experiment, the fringe of width $0.05mm$ is observed on a screen at a distance of $1.5m$ from the source . When a convex lens is placed between the source and the screen, for two positions of the lens image of interfering sources are produced on the screen. The separation between the two images being $0.04$ and $0.01mm$, respectively. The wavelength of light used is

physics wave optics difference between interference and diffraction explaining wave phenomena diffraction

  1. $6.67nm$

  2. $0.667nm$

  3. $667nm$

  4. $667A^o$

Show answer
Correct Option: B
Explanation:

Distance between slits, $d=\sqrt{d _{1}d _{2}}=0..02mm$

Fringe width, $\beta=\dfrac{D\lambda}{d}$
hence $\lambda=\dfrac{\beta d}{D}=0.667nm$

How can resolving power of the instrument be increased?

physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.

Show answer
Correct Option: A,B
Explanation:

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

The ability of an optical instruments to show the images of two adjacent point objects as separate is called :

physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these

Show answer
Correct Option: C
Explanation:

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

Two lenses of focal lengths $+ 100 cm$ and $+ 5 cm$ are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at $\displaystyle \infty $)

physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

  1. $95 cm$

  2. $100 cm$

  3. $105 cm$

  4. $500 cm$

Show answer
Correct Option: C
Explanation:

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$

In optical instruments, the lenses are used to form images by :

physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

  1. Reflection

  2. Refraction

  3. Dispersion

  4. Scattering

Show answer
Correct Option: B
Explanation:

In optical instruments, the lenses are used to form images by Refraction.