283436
In a Young's double slit experiment, distance between the two slits is \(1 \mathrm{~mm}\) and the distance between the screen and the two slits is \(1 \mathrm{~m}\). If the fringe width on the screen is \(0.06 \mathrm{~cm}\) then the wavelength of light is
283438
In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance \(D\) of the screen from the slits would be made
1 \(2 \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{\mathrm{D}}{2}\)
4 \(\frac{D}{4}\)
Explanation:
: Given, \(d^{\prime}=2 d\) \(\beta^{\prime}=\beta\) Young's double-slit experiment the fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\mathrm{d}=\) distance between slits \(\mathrm{D}=\) distance between slits and screen \(\lambda=\) wavelength \(\frac{\lambda \mathrm{D}^{\prime}}{\mathrm{d}^{\prime}}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \quad\left(\begin{array}{c}\mathrm{D}^{\prime}= \\ \begin{array}{c}\text { New value of distance of } \\ \text { screen from double slit }\end{array}\end{array}\right)\) \(\frac{D^{\prime}}{2 d}=\frac{D}{d}\) \(D^{\prime}=2 D\)
UPSEE - 2010
WAVE OPTICS
283439
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
1 \(\frac{\pi}{4}\) radian
2 \(\frac{\pi}{2}\) radian
3 \(\pi\) radian
4 \(\frac{\pi}{8}\) radian
Explanation:
: For the first minima at \(\mathrm{P}\) a \(\sin \theta=\lambda\) \(\sin \theta=\frac{\Delta x}{a / 2}\) \(\Delta x=(a / 2) \sin \theta\) Phase difference \((\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}\) \(=\frac{2 \pi}{\lambda} \times \frac{\mathrm{a}}{2} \sin \theta \quad(\because \mathrm{a} \sin \theta=\lambda)\) \(=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) \(=\pi \text { radian }\)
AIPMT - 2015
WAVE OPTICS
283440
In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle \(30^{\circ}\) when of wavelength \(5000 \AA\) is incident on the slit. The first secondary maxima is observed at an angle of
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{1}{2}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
Explanation:
: For \(\mathrm{n}^{\text {th }}\) minima \(\text { a } \sin \theta=\mathrm{n} \lambda\) \(\mathrm{a}=\text { slit width }\) At \(\mathrm{n}=1, \theta=30^{\circ}\) \(\mathrm{a} \sin 30^{\circ}=1 \times \lambda\) \(\mathrm{a}=2 \lambda\) For \(\mathrm{n}^{\text {th }}\) secondary maxima- \(a \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda\) For first secondary maxima- \(\text { at } \mathrm{n}=1\) a \(\sin \theta=\frac{3 \lambda}{2}\) \(2 \lambda \times \sin \theta=\frac{3 \lambda}{2}\) \(\sin \theta=\frac{3}{4}\) \(\theta=\sin ^{-1}\left(\frac{3}{4}\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVE OPTICS
283436
In a Young's double slit experiment, distance between the two slits is \(1 \mathrm{~mm}\) and the distance between the screen and the two slits is \(1 \mathrm{~m}\). If the fringe width on the screen is \(0.06 \mathrm{~cm}\) then the wavelength of light is
283438
In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance \(D\) of the screen from the slits would be made
1 \(2 \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{\mathrm{D}}{2}\)
4 \(\frac{D}{4}\)
Explanation:
: Given, \(d^{\prime}=2 d\) \(\beta^{\prime}=\beta\) Young's double-slit experiment the fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\mathrm{d}=\) distance between slits \(\mathrm{D}=\) distance between slits and screen \(\lambda=\) wavelength \(\frac{\lambda \mathrm{D}^{\prime}}{\mathrm{d}^{\prime}}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \quad\left(\begin{array}{c}\mathrm{D}^{\prime}= \\ \begin{array}{c}\text { New value of distance of } \\ \text { screen from double slit }\end{array}\end{array}\right)\) \(\frac{D^{\prime}}{2 d}=\frac{D}{d}\) \(D^{\prime}=2 D\)
UPSEE - 2010
WAVE OPTICS
283439
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
1 \(\frac{\pi}{4}\) radian
2 \(\frac{\pi}{2}\) radian
3 \(\pi\) radian
4 \(\frac{\pi}{8}\) radian
Explanation:
: For the first minima at \(\mathrm{P}\) a \(\sin \theta=\lambda\) \(\sin \theta=\frac{\Delta x}{a / 2}\) \(\Delta x=(a / 2) \sin \theta\) Phase difference \((\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}\) \(=\frac{2 \pi}{\lambda} \times \frac{\mathrm{a}}{2} \sin \theta \quad(\because \mathrm{a} \sin \theta=\lambda)\) \(=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) \(=\pi \text { radian }\)
AIPMT - 2015
WAVE OPTICS
283440
In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle \(30^{\circ}\) when of wavelength \(5000 \AA\) is incident on the slit. The first secondary maxima is observed at an angle of
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{1}{2}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
Explanation:
: For \(\mathrm{n}^{\text {th }}\) minima \(\text { a } \sin \theta=\mathrm{n} \lambda\) \(\mathrm{a}=\text { slit width }\) At \(\mathrm{n}=1, \theta=30^{\circ}\) \(\mathrm{a} \sin 30^{\circ}=1 \times \lambda\) \(\mathrm{a}=2 \lambda\) For \(\mathrm{n}^{\text {th }}\) secondary maxima- \(a \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda\) For first secondary maxima- \(\text { at } \mathrm{n}=1\) a \(\sin \theta=\frac{3 \lambda}{2}\) \(2 \lambda \times \sin \theta=\frac{3 \lambda}{2}\) \(\sin \theta=\frac{3}{4}\) \(\theta=\sin ^{-1}\left(\frac{3}{4}\right)\)
283436
In a Young's double slit experiment, distance between the two slits is \(1 \mathrm{~mm}\) and the distance between the screen and the two slits is \(1 \mathrm{~m}\). If the fringe width on the screen is \(0.06 \mathrm{~cm}\) then the wavelength of light is
283438
In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance \(D\) of the screen from the slits would be made
1 \(2 \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{\mathrm{D}}{2}\)
4 \(\frac{D}{4}\)
Explanation:
: Given, \(d^{\prime}=2 d\) \(\beta^{\prime}=\beta\) Young's double-slit experiment the fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\mathrm{d}=\) distance between slits \(\mathrm{D}=\) distance between slits and screen \(\lambda=\) wavelength \(\frac{\lambda \mathrm{D}^{\prime}}{\mathrm{d}^{\prime}}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \quad\left(\begin{array}{c}\mathrm{D}^{\prime}= \\ \begin{array}{c}\text { New value of distance of } \\ \text { screen from double slit }\end{array}\end{array}\right)\) \(\frac{D^{\prime}}{2 d}=\frac{D}{d}\) \(D^{\prime}=2 D\)
UPSEE - 2010
WAVE OPTICS
283439
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
1 \(\frac{\pi}{4}\) radian
2 \(\frac{\pi}{2}\) radian
3 \(\pi\) radian
4 \(\frac{\pi}{8}\) radian
Explanation:
: For the first minima at \(\mathrm{P}\) a \(\sin \theta=\lambda\) \(\sin \theta=\frac{\Delta x}{a / 2}\) \(\Delta x=(a / 2) \sin \theta\) Phase difference \((\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}\) \(=\frac{2 \pi}{\lambda} \times \frac{\mathrm{a}}{2} \sin \theta \quad(\because \mathrm{a} \sin \theta=\lambda)\) \(=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) \(=\pi \text { radian }\)
AIPMT - 2015
WAVE OPTICS
283440
In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle \(30^{\circ}\) when of wavelength \(5000 \AA\) is incident on the slit. The first secondary maxima is observed at an angle of
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{1}{2}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
Explanation:
: For \(\mathrm{n}^{\text {th }}\) minima \(\text { a } \sin \theta=\mathrm{n} \lambda\) \(\mathrm{a}=\text { slit width }\) At \(\mathrm{n}=1, \theta=30^{\circ}\) \(\mathrm{a} \sin 30^{\circ}=1 \times \lambda\) \(\mathrm{a}=2 \lambda\) For \(\mathrm{n}^{\text {th }}\) secondary maxima- \(a \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda\) For first secondary maxima- \(\text { at } \mathrm{n}=1\) a \(\sin \theta=\frac{3 \lambda}{2}\) \(2 \lambda \times \sin \theta=\frac{3 \lambda}{2}\) \(\sin \theta=\frac{3}{4}\) \(\theta=\sin ^{-1}\left(\frac{3}{4}\right)\)
283436
In a Young's double slit experiment, distance between the two slits is \(1 \mathrm{~mm}\) and the distance between the screen and the two slits is \(1 \mathrm{~m}\). If the fringe width on the screen is \(0.06 \mathrm{~cm}\) then the wavelength of light is
283438
In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance \(D\) of the screen from the slits would be made
1 \(2 \mathrm{D}\)
2 \(\mathrm{D}\)
3 \(\frac{\mathrm{D}}{2}\)
4 \(\frac{D}{4}\)
Explanation:
: Given, \(d^{\prime}=2 d\) \(\beta^{\prime}=\beta\) Young's double-slit experiment the fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\) Where, \(\mathrm{d}=\) distance between slits \(\mathrm{D}=\) distance between slits and screen \(\lambda=\) wavelength \(\frac{\lambda \mathrm{D}^{\prime}}{\mathrm{d}^{\prime}}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \quad\left(\begin{array}{c}\mathrm{D}^{\prime}= \\ \begin{array}{c}\text { New value of distance of } \\ \text { screen from double slit }\end{array}\end{array}\right)\) \(\frac{D^{\prime}}{2 d}=\frac{D}{d}\) \(D^{\prime}=2 D\)
UPSEE - 2010
WAVE OPTICS
283439
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is
1 \(\frac{\pi}{4}\) radian
2 \(\frac{\pi}{2}\) radian
3 \(\pi\) radian
4 \(\frac{\pi}{8}\) radian
Explanation:
: For the first minima at \(\mathrm{P}\) a \(\sin \theta=\lambda\) \(\sin \theta=\frac{\Delta x}{a / 2}\) \(\Delta x=(a / 2) \sin \theta\) Phase difference \((\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}\) \(=\frac{2 \pi}{\lambda} \times \frac{\mathrm{a}}{2} \sin \theta \quad(\because \mathrm{a} \sin \theta=\lambda)\) \(=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) \(=\pi \text { radian }\)
AIPMT - 2015
WAVE OPTICS
283440
In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle \(30^{\circ}\) when of wavelength \(5000 \AA\) is incident on the slit. The first secondary maxima is observed at an angle of
1 \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2 \(\sin ^{-1}\left(\frac{1}{2}\right)\)
3 \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4 \(\sin ^{-1}\left(\frac{1}{4}\right)\)
Explanation:
: For \(\mathrm{n}^{\text {th }}\) minima \(\text { a } \sin \theta=\mathrm{n} \lambda\) \(\mathrm{a}=\text { slit width }\) At \(\mathrm{n}=1, \theta=30^{\circ}\) \(\mathrm{a} \sin 30^{\circ}=1 \times \lambda\) \(\mathrm{a}=2 \lambda\) For \(\mathrm{n}^{\text {th }}\) secondary maxima- \(a \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda\) For first secondary maxima- \(\text { at } \mathrm{n}=1\) a \(\sin \theta=\frac{3 \lambda}{2}\) \(2 \lambda \times \sin \theta=\frac{3 \lambda}{2}\) \(\sin \theta=\frac{3}{4}\) \(\theta=\sin ^{-1}\left(\frac{3}{4}\right)\)
