283211
In young's double slit experiment, to increase the fringe width
1 the wavelength of the source is increased
2 the source is moved towards the slit
3 the source is move away from the slit
4 the slit separation is increased
5 the screen is move towards the slit
Explanation:
: By Young's double slit experiment fringe width, \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) According to equation the fringe width is directly proportional to \(\lambda\). So, as we increases the fringe width the wavelength of the source is increased.
Kerala CEE- 2014
WAVE OPTICS
283220
If path difference becomes \((2 n-1) \frac{\lambda}{2}\) then:
1 white fringe is formed
2 bright fringe is formed
3 uniform illumination is obtained
4 dark fringe is formed
Explanation:
: As the path difference is odd multiple of \(\frac{\lambda}{2}\) Here, path difference \(=(2 n-1) \frac{\lambda}{2}\) Which is an odd multiple of \(\frac{\lambda}{2}\) Hence, destructive interference takes place and dark fringe is formed.
JCECE-2006
WAVE OPTICS
283221
The fringe width for red light is approximately how many times that for violet light in Young's slit experiment?
1 2 times
2 3 times
3 Equal
4 \(1 / 2\) times
Explanation:
: We know, For red color \(\left(\lambda_{\mathrm{r}}\right)=620-750 \mathrm{~nm}\) Violet color \(\left(\lambda_{\mathrm{v}}\right)=380-450 \mathrm{~nm}\) \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)Then, \(\quad \lambda_{\text {red }} \cong 2 \lambda_{\text {violet }}\) So, \(\quad \beta_{\text {red }} \cong 2 \beta_{\text {violet }}\)
JCECE-2006
WAVE OPTICS
283231
Interference occurs in which of the following waves?
1 Transverse
2 Electromagnetic
3 Longitudinal
4 All of these
Explanation:
: Interference can occur in transverse waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in longitudinal waves when the compression and refraction of the waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in electromagnetic waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Hence, interference can occur in transverse wave, longitudinal wave and electromagnetic wave.
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WAVE OPTICS
283211
In young's double slit experiment, to increase the fringe width
1 the wavelength of the source is increased
2 the source is moved towards the slit
3 the source is move away from the slit
4 the slit separation is increased
5 the screen is move towards the slit
Explanation:
: By Young's double slit experiment fringe width, \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) According to equation the fringe width is directly proportional to \(\lambda\). So, as we increases the fringe width the wavelength of the source is increased.
Kerala CEE- 2014
WAVE OPTICS
283220
If path difference becomes \((2 n-1) \frac{\lambda}{2}\) then:
1 white fringe is formed
2 bright fringe is formed
3 uniform illumination is obtained
4 dark fringe is formed
Explanation:
: As the path difference is odd multiple of \(\frac{\lambda}{2}\) Here, path difference \(=(2 n-1) \frac{\lambda}{2}\) Which is an odd multiple of \(\frac{\lambda}{2}\) Hence, destructive interference takes place and dark fringe is formed.
JCECE-2006
WAVE OPTICS
283221
The fringe width for red light is approximately how many times that for violet light in Young's slit experiment?
1 2 times
2 3 times
3 Equal
4 \(1 / 2\) times
Explanation:
: We know, For red color \(\left(\lambda_{\mathrm{r}}\right)=620-750 \mathrm{~nm}\) Violet color \(\left(\lambda_{\mathrm{v}}\right)=380-450 \mathrm{~nm}\) \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)Then, \(\quad \lambda_{\text {red }} \cong 2 \lambda_{\text {violet }}\) So, \(\quad \beta_{\text {red }} \cong 2 \beta_{\text {violet }}\)
JCECE-2006
WAVE OPTICS
283231
Interference occurs in which of the following waves?
1 Transverse
2 Electromagnetic
3 Longitudinal
4 All of these
Explanation:
: Interference can occur in transverse waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in longitudinal waves when the compression and refraction of the waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in electromagnetic waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Hence, interference can occur in transverse wave, longitudinal wave and electromagnetic wave.
283211
In young's double slit experiment, to increase the fringe width
1 the wavelength of the source is increased
2 the source is moved towards the slit
3 the source is move away from the slit
4 the slit separation is increased
5 the screen is move towards the slit
Explanation:
: By Young's double slit experiment fringe width, \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) According to equation the fringe width is directly proportional to \(\lambda\). So, as we increases the fringe width the wavelength of the source is increased.
Kerala CEE- 2014
WAVE OPTICS
283220
If path difference becomes \((2 n-1) \frac{\lambda}{2}\) then:
1 white fringe is formed
2 bright fringe is formed
3 uniform illumination is obtained
4 dark fringe is formed
Explanation:
: As the path difference is odd multiple of \(\frac{\lambda}{2}\) Here, path difference \(=(2 n-1) \frac{\lambda}{2}\) Which is an odd multiple of \(\frac{\lambda}{2}\) Hence, destructive interference takes place and dark fringe is formed.
JCECE-2006
WAVE OPTICS
283221
The fringe width for red light is approximately how many times that for violet light in Young's slit experiment?
1 2 times
2 3 times
3 Equal
4 \(1 / 2\) times
Explanation:
: We know, For red color \(\left(\lambda_{\mathrm{r}}\right)=620-750 \mathrm{~nm}\) Violet color \(\left(\lambda_{\mathrm{v}}\right)=380-450 \mathrm{~nm}\) \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)Then, \(\quad \lambda_{\text {red }} \cong 2 \lambda_{\text {violet }}\) So, \(\quad \beta_{\text {red }} \cong 2 \beta_{\text {violet }}\)
JCECE-2006
WAVE OPTICS
283231
Interference occurs in which of the following waves?
1 Transverse
2 Electromagnetic
3 Longitudinal
4 All of these
Explanation:
: Interference can occur in transverse waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in longitudinal waves when the compression and refraction of the waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in electromagnetic waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Hence, interference can occur in transverse wave, longitudinal wave and electromagnetic wave.
283211
In young's double slit experiment, to increase the fringe width
1 the wavelength of the source is increased
2 the source is moved towards the slit
3 the source is move away from the slit
4 the slit separation is increased
5 the screen is move towards the slit
Explanation:
: By Young's double slit experiment fringe width, \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) According to equation the fringe width is directly proportional to \(\lambda\). So, as we increases the fringe width the wavelength of the source is increased.
Kerala CEE- 2014
WAVE OPTICS
283220
If path difference becomes \((2 n-1) \frac{\lambda}{2}\) then:
1 white fringe is formed
2 bright fringe is formed
3 uniform illumination is obtained
4 dark fringe is formed
Explanation:
: As the path difference is odd multiple of \(\frac{\lambda}{2}\) Here, path difference \(=(2 n-1) \frac{\lambda}{2}\) Which is an odd multiple of \(\frac{\lambda}{2}\) Hence, destructive interference takes place and dark fringe is formed.
JCECE-2006
WAVE OPTICS
283221
The fringe width for red light is approximately how many times that for violet light in Young's slit experiment?
1 2 times
2 3 times
3 Equal
4 \(1 / 2\) times
Explanation:
: We know, For red color \(\left(\lambda_{\mathrm{r}}\right)=620-750 \mathrm{~nm}\) Violet color \(\left(\lambda_{\mathrm{v}}\right)=380-450 \mathrm{~nm}\) \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)Then, \(\quad \lambda_{\text {red }} \cong 2 \lambda_{\text {violet }}\) So, \(\quad \beta_{\text {red }} \cong 2 \beta_{\text {violet }}\)
JCECE-2006
WAVE OPTICS
283231
Interference occurs in which of the following waves?
1 Transverse
2 Electromagnetic
3 Longitudinal
4 All of these
Explanation:
: Interference can occur in transverse waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in longitudinal waves when the compression and refraction of the waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Interference can occur in electromagnetic waves when two waves meet each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves. Hence, interference can occur in transverse wave, longitudinal wave and electromagnetic wave.