367709
Assertion : If the slit width is equal to wavelength in diffraction then the entire screen is bright. Reason : Angular position of \({1^{st}}\) minima is \(\sin \theta = \frac{{2\lambda }}{a}\)
1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
3 Assertion is true but the Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Angular position of 1st minima is \(\sin \theta = \frac{\lambda }{a}\,{\rm{if}}\,\lambda = a\) \( \Rightarrow \theta = {90^o}\) i.e., 1st minima will not appear on the screen. So the entire screen is bright. So option (3) is correct.
PHXII10:WAVE OPTICS
367710
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400\,nm\), the first minimum is formed at an angle of \(30^\circ \) . The direction \(\theta \) of the first secondary maximum is given by
1 \({\sin ^{ - 1}}{\rm{ }}(1/4)\)
2 \({\tan ^{ - 1{\rm{ }}}}(2/3){\rm{ }}\)
3 \({\sin ^{ - 1{\rm{ }}}}(2{\rm{/}}3)\)
4 \({\sin ^{ - 1{\rm{ }}}}(3/4)\)
Explanation:
For Fraunhofer diffraction at single slit, the equation for secondary maxima is given by \(\sin \theta = \frac{{(2n + 1)\lambda }}{{2a}}\) For the first minimum, \(a\sin \theta = \lambda \Rightarrow \frac{\lambda }{a} = \sin \theta = \sin 30^\circ = \frac{1}{2}\) \(\therefore \) For first secondary maximum \(\sin \theta = \frac{{(2 \times 1 + 1)}}{2} \times \frac{1}{2} = \frac{3}{4}\) \(\therefore \theta = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)\)
KCET - 2008
PHXII10:WAVE OPTICS
367711
First diffraction minima due to single slit diffraction is at \(\theta = 30^\circ \) for wavelength \(6000\mathop A\limits^ \circ \). The width of slit is
367712
Light of wavelength \(600\;nm\) is incident normally on a slit of width \(0.2\;mm\) . The angular width of central maxima in the diffraction pattern is (measured from minimum ot minimum)
367709
Assertion : If the slit width is equal to wavelength in diffraction then the entire screen is bright. Reason : Angular position of \({1^{st}}\) minima is \(\sin \theta = \frac{{2\lambda }}{a}\)
1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
3 Assertion is true but the Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Angular position of 1st minima is \(\sin \theta = \frac{\lambda }{a}\,{\rm{if}}\,\lambda = a\) \( \Rightarrow \theta = {90^o}\) i.e., 1st minima will not appear on the screen. So the entire screen is bright. So option (3) is correct.
PHXII10:WAVE OPTICS
367710
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400\,nm\), the first minimum is formed at an angle of \(30^\circ \) . The direction \(\theta \) of the first secondary maximum is given by
1 \({\sin ^{ - 1}}{\rm{ }}(1/4)\)
2 \({\tan ^{ - 1{\rm{ }}}}(2/3){\rm{ }}\)
3 \({\sin ^{ - 1{\rm{ }}}}(2{\rm{/}}3)\)
4 \({\sin ^{ - 1{\rm{ }}}}(3/4)\)
Explanation:
For Fraunhofer diffraction at single slit, the equation for secondary maxima is given by \(\sin \theta = \frac{{(2n + 1)\lambda }}{{2a}}\) For the first minimum, \(a\sin \theta = \lambda \Rightarrow \frac{\lambda }{a} = \sin \theta = \sin 30^\circ = \frac{1}{2}\) \(\therefore \) For first secondary maximum \(\sin \theta = \frac{{(2 \times 1 + 1)}}{2} \times \frac{1}{2} = \frac{3}{4}\) \(\therefore \theta = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)\)
KCET - 2008
PHXII10:WAVE OPTICS
367711
First diffraction minima due to single slit diffraction is at \(\theta = 30^\circ \) for wavelength \(6000\mathop A\limits^ \circ \). The width of slit is
367712
Light of wavelength \(600\;nm\) is incident normally on a slit of width \(0.2\;mm\) . The angular width of central maxima in the diffraction pattern is (measured from minimum ot minimum)
367709
Assertion : If the slit width is equal to wavelength in diffraction then the entire screen is bright. Reason : Angular position of \({1^{st}}\) minima is \(\sin \theta = \frac{{2\lambda }}{a}\)
1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
3 Assertion is true but the Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Angular position of 1st minima is \(\sin \theta = \frac{\lambda }{a}\,{\rm{if}}\,\lambda = a\) \( \Rightarrow \theta = {90^o}\) i.e., 1st minima will not appear on the screen. So the entire screen is bright. So option (3) is correct.
PHXII10:WAVE OPTICS
367710
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400\,nm\), the first minimum is formed at an angle of \(30^\circ \) . The direction \(\theta \) of the first secondary maximum is given by
1 \({\sin ^{ - 1}}{\rm{ }}(1/4)\)
2 \({\tan ^{ - 1{\rm{ }}}}(2/3){\rm{ }}\)
3 \({\sin ^{ - 1{\rm{ }}}}(2{\rm{/}}3)\)
4 \({\sin ^{ - 1{\rm{ }}}}(3/4)\)
Explanation:
For Fraunhofer diffraction at single slit, the equation for secondary maxima is given by \(\sin \theta = \frac{{(2n + 1)\lambda }}{{2a}}\) For the first minimum, \(a\sin \theta = \lambda \Rightarrow \frac{\lambda }{a} = \sin \theta = \sin 30^\circ = \frac{1}{2}\) \(\therefore \) For first secondary maximum \(\sin \theta = \frac{{(2 \times 1 + 1)}}{2} \times \frac{1}{2} = \frac{3}{4}\) \(\therefore \theta = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)\)
KCET - 2008
PHXII10:WAVE OPTICS
367711
First diffraction minima due to single slit diffraction is at \(\theta = 30^\circ \) for wavelength \(6000\mathop A\limits^ \circ \). The width of slit is
367712
Light of wavelength \(600\;nm\) is incident normally on a slit of width \(0.2\;mm\) . The angular width of central maxima in the diffraction pattern is (measured from minimum ot minimum)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII10:WAVE OPTICS
367709
Assertion : If the slit width is equal to wavelength in diffraction then the entire screen is bright. Reason : Angular position of \({1^{st}}\) minima is \(\sin \theta = \frac{{2\lambda }}{a}\)
1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.
2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
3 Assertion is true but the Reason is false.
4 Assertion is false but Reason is true.
Explanation:
Angular position of 1st minima is \(\sin \theta = \frac{\lambda }{a}\,{\rm{if}}\,\lambda = a\) \( \Rightarrow \theta = {90^o}\) i.e., 1st minima will not appear on the screen. So the entire screen is bright. So option (3) is correct.
PHXII10:WAVE OPTICS
367710
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength \(400\,nm\), the first minimum is formed at an angle of \(30^\circ \) . The direction \(\theta \) of the first secondary maximum is given by
1 \({\sin ^{ - 1}}{\rm{ }}(1/4)\)
2 \({\tan ^{ - 1{\rm{ }}}}(2/3){\rm{ }}\)
3 \({\sin ^{ - 1{\rm{ }}}}(2{\rm{/}}3)\)
4 \({\sin ^{ - 1{\rm{ }}}}(3/4)\)
Explanation:
For Fraunhofer diffraction at single slit, the equation for secondary maxima is given by \(\sin \theta = \frac{{(2n + 1)\lambda }}{{2a}}\) For the first minimum, \(a\sin \theta = \lambda \Rightarrow \frac{\lambda }{a} = \sin \theta = \sin 30^\circ = \frac{1}{2}\) \(\therefore \) For first secondary maximum \(\sin \theta = \frac{{(2 \times 1 + 1)}}{2} \times \frac{1}{2} = \frac{3}{4}\) \(\therefore \theta = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)\)
KCET - 2008
PHXII10:WAVE OPTICS
367711
First diffraction minima due to single slit diffraction is at \(\theta = 30^\circ \) for wavelength \(6000\mathop A\limits^ \circ \). The width of slit is
367712
Light of wavelength \(600\;nm\) is incident normally on a slit of width \(0.2\;mm\) . The angular width of central maxima in the diffraction pattern is (measured from minimum ot minimum)
