368051
In double-slit experiment using light of wavelength \(600\;nm\) the angular width of a fringe formed on a distant screen is \(0.1^\circ \). What is the spacing between the two slits?
1 \(3.03 \times {10^{ - 4}}{\rm{ }}m\)
2 \(3.44 \times {10^{ - 4}}{\rm{ }}m\)
3 \(2.68 \times {10^{ - 4}}{\rm{ }}m\)
4 \(4.03 \times {10^{ - 4}}{\rm{ }}m\)
Explanation:
Given, \(\lambda = 600\,nm = 600 \times {10^{ - 9}}m\) Angular width of fringe \(\theta = 0.1^\circ = \frac{{0.1\pi }}{{180}}\,rad\) Using the formula \(\theta = \frac{\lambda }{d}\) Spacing between the slits \(d = \frac{\lambda }{\theta } = \frac{{600 \times {{10}^{ - 9}} \times 180}}{{0.1 \times \pi }}\) \(d = 3.44 \times {10^{ - 4}}\,m\)
NCERT Exemplar
PHXII10:WAVE OPTICS
368052
In Young’s double slit experiment, the angular width of the fringes is \({0.20^ \circ }\) for the sodium light \(\lambda = 5890\mathop A\limits^ \circ \)]. In order to increase the angular width of the fringes by \(10\% \) the necessary change in wavelength is:
1 Zero
2 increased by \(6479{\rm{ }}\mathop A\limits^ \circ \)
3 decreased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
4 increased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
Explanation:
Let \(\lambda \) be the wavelength of monochromatic light, then the angular fringe width is \(\theta = \frac{\lambda }{d}\) Where, \(d\) is distance between coherent sources. Given, \(\frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}}\) \(\frac{{\Delta \lambda }}{\lambda } = \frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}} = 0.1\) \( \Rightarrow {\rm{ }}\Delta \lambda = 0.1\lambda = 0.1 \times 5890{\rm{ }}\mathop A\limits^ \circ = 589\mathop {{\rm{ }}A}\limits^ \circ \)
PHXII10:WAVE OPTICS
368053
In young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
368054
In Young’s double slit experiment green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Moving the screen away from the slits
2 Using blue light instead of green light
3 Using red light instead of green light
4 Reducing the separation between the slits
Explanation:
The distance between the fringe or the fringe width is given by \(\beta = \frac{{D\lambda }}{d}\) where, \(\lambda = \) wavelength of light used \(D\) = distance of slits from screen and \(d\) = separation between slits So, to reduce the space between fringes should be reduced. As, blue light has shorter wavelength than that of green light. So, by using blue light instead of green light, the fringes become more closely.
368051
In double-slit experiment using light of wavelength \(600\;nm\) the angular width of a fringe formed on a distant screen is \(0.1^\circ \). What is the spacing between the two slits?
1 \(3.03 \times {10^{ - 4}}{\rm{ }}m\)
2 \(3.44 \times {10^{ - 4}}{\rm{ }}m\)
3 \(2.68 \times {10^{ - 4}}{\rm{ }}m\)
4 \(4.03 \times {10^{ - 4}}{\rm{ }}m\)
Explanation:
Given, \(\lambda = 600\,nm = 600 \times {10^{ - 9}}m\) Angular width of fringe \(\theta = 0.1^\circ = \frac{{0.1\pi }}{{180}}\,rad\) Using the formula \(\theta = \frac{\lambda }{d}\) Spacing between the slits \(d = \frac{\lambda }{\theta } = \frac{{600 \times {{10}^{ - 9}} \times 180}}{{0.1 \times \pi }}\) \(d = 3.44 \times {10^{ - 4}}\,m\)
NCERT Exemplar
PHXII10:WAVE OPTICS
368052
In Young’s double slit experiment, the angular width of the fringes is \({0.20^ \circ }\) for the sodium light \(\lambda = 5890\mathop A\limits^ \circ \)]. In order to increase the angular width of the fringes by \(10\% \) the necessary change in wavelength is:
1 Zero
2 increased by \(6479{\rm{ }}\mathop A\limits^ \circ \)
3 decreased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
4 increased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
Explanation:
Let \(\lambda \) be the wavelength of monochromatic light, then the angular fringe width is \(\theta = \frac{\lambda }{d}\) Where, \(d\) is distance between coherent sources. Given, \(\frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}}\) \(\frac{{\Delta \lambda }}{\lambda } = \frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}} = 0.1\) \( \Rightarrow {\rm{ }}\Delta \lambda = 0.1\lambda = 0.1 \times 5890{\rm{ }}\mathop A\limits^ \circ = 589\mathop {{\rm{ }}A}\limits^ \circ \)
PHXII10:WAVE OPTICS
368053
In young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
368054
In Young’s double slit experiment green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Moving the screen away from the slits
2 Using blue light instead of green light
3 Using red light instead of green light
4 Reducing the separation between the slits
Explanation:
The distance between the fringe or the fringe width is given by \(\beta = \frac{{D\lambda }}{d}\) where, \(\lambda = \) wavelength of light used \(D\) = distance of slits from screen and \(d\) = separation between slits So, to reduce the space between fringes should be reduced. As, blue light has shorter wavelength than that of green light. So, by using blue light instead of green light, the fringes become more closely.
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PHXII10:WAVE OPTICS
368051
In double-slit experiment using light of wavelength \(600\;nm\) the angular width of a fringe formed on a distant screen is \(0.1^\circ \). What is the spacing between the two slits?
1 \(3.03 \times {10^{ - 4}}{\rm{ }}m\)
2 \(3.44 \times {10^{ - 4}}{\rm{ }}m\)
3 \(2.68 \times {10^{ - 4}}{\rm{ }}m\)
4 \(4.03 \times {10^{ - 4}}{\rm{ }}m\)
Explanation:
Given, \(\lambda = 600\,nm = 600 \times {10^{ - 9}}m\) Angular width of fringe \(\theta = 0.1^\circ = \frac{{0.1\pi }}{{180}}\,rad\) Using the formula \(\theta = \frac{\lambda }{d}\) Spacing between the slits \(d = \frac{\lambda }{\theta } = \frac{{600 \times {{10}^{ - 9}} \times 180}}{{0.1 \times \pi }}\) \(d = 3.44 \times {10^{ - 4}}\,m\)
NCERT Exemplar
PHXII10:WAVE OPTICS
368052
In Young’s double slit experiment, the angular width of the fringes is \({0.20^ \circ }\) for the sodium light \(\lambda = 5890\mathop A\limits^ \circ \)]. In order to increase the angular width of the fringes by \(10\% \) the necessary change in wavelength is:
1 Zero
2 increased by \(6479{\rm{ }}\mathop A\limits^ \circ \)
3 decreased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
4 increased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
Explanation:
Let \(\lambda \) be the wavelength of monochromatic light, then the angular fringe width is \(\theta = \frac{\lambda }{d}\) Where, \(d\) is distance between coherent sources. Given, \(\frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}}\) \(\frac{{\Delta \lambda }}{\lambda } = \frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}} = 0.1\) \( \Rightarrow {\rm{ }}\Delta \lambda = 0.1\lambda = 0.1 \times 5890{\rm{ }}\mathop A\limits^ \circ = 589\mathop {{\rm{ }}A}\limits^ \circ \)
PHXII10:WAVE OPTICS
368053
In young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
368054
In Young’s double slit experiment green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Moving the screen away from the slits
2 Using blue light instead of green light
3 Using red light instead of green light
4 Reducing the separation between the slits
Explanation:
The distance between the fringe or the fringe width is given by \(\beta = \frac{{D\lambda }}{d}\) where, \(\lambda = \) wavelength of light used \(D\) = distance of slits from screen and \(d\) = separation between slits So, to reduce the space between fringes should be reduced. As, blue light has shorter wavelength than that of green light. So, by using blue light instead of green light, the fringes become more closely.
368051
In double-slit experiment using light of wavelength \(600\;nm\) the angular width of a fringe formed on a distant screen is \(0.1^\circ \). What is the spacing between the two slits?
1 \(3.03 \times {10^{ - 4}}{\rm{ }}m\)
2 \(3.44 \times {10^{ - 4}}{\rm{ }}m\)
3 \(2.68 \times {10^{ - 4}}{\rm{ }}m\)
4 \(4.03 \times {10^{ - 4}}{\rm{ }}m\)
Explanation:
Given, \(\lambda = 600\,nm = 600 \times {10^{ - 9}}m\) Angular width of fringe \(\theta = 0.1^\circ = \frac{{0.1\pi }}{{180}}\,rad\) Using the formula \(\theta = \frac{\lambda }{d}\) Spacing between the slits \(d = \frac{\lambda }{\theta } = \frac{{600 \times {{10}^{ - 9}} \times 180}}{{0.1 \times \pi }}\) \(d = 3.44 \times {10^{ - 4}}\,m\)
NCERT Exemplar
PHXII10:WAVE OPTICS
368052
In Young’s double slit experiment, the angular width of the fringes is \({0.20^ \circ }\) for the sodium light \(\lambda = 5890\mathop A\limits^ \circ \)]. In order to increase the angular width of the fringes by \(10\% \) the necessary change in wavelength is:
1 Zero
2 increased by \(6479{\rm{ }}\mathop A\limits^ \circ \)
3 decreased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
4 increased by \(589\mathop {{\rm{ }}A}\limits^ \circ {\rm{ }}\)
Explanation:
Let \(\lambda \) be the wavelength of monochromatic light, then the angular fringe width is \(\theta = \frac{\lambda }{d}\) Where, \(d\) is distance between coherent sources. Given, \(\frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}}\) \(\frac{{\Delta \lambda }}{\lambda } = \frac{{\Delta \theta }}{\theta } = \frac{{10}}{{100}} = 0.1\) \( \Rightarrow {\rm{ }}\Delta \lambda = 0.1\lambda = 0.1 \times 5890{\rm{ }}\mathop A\limits^ \circ = 589\mathop {{\rm{ }}A}\limits^ \circ \)
PHXII10:WAVE OPTICS
368053
In young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
368054
In Young’s double slit experiment green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced?
1 Moving the screen away from the slits
2 Using blue light instead of green light
3 Using red light instead of green light
4 Reducing the separation between the slits
Explanation:
The distance between the fringe or the fringe width is given by \(\beta = \frac{{D\lambda }}{d}\) where, \(\lambda = \) wavelength of light used \(D\) = distance of slits from screen and \(d\) = separation between slits So, to reduce the space between fringes should be reduced. As, blue light has shorter wavelength than that of green light. So, by using blue light instead of green light, the fringes become more closely.