283083
In an electron microscope the accelerating voltage is increased from \(20 \mathrm{kV}\) to \(80 \mathrm{kV}\), the resolving power of the microscope will become
1 \(2 \mathrm{R}\)
2 \(\frac{R}{2}\)
3 \(4 \mathrm{R}\)
4 \(3 \mathrm{R}\)
Explanation:
: Wavelength and accelerating voltage for an electron, \(\lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{V}=\text { potential applied })\) \(\because\) Resolving power \(\propto \frac{1}{\lambda}\) \(\therefore\) Resolving power \(\propto \sqrt{\mathrm{V}}\) Now, if potential used is increased 4 times, resolving power will be increased 2 times. So, if resolving power earlier is R. then it becomes \(2 \mathrm{R}\).
AIIMS-2000
WAVE OPTICS
283085
According to Hubble's law, the red-shift \((\mathrm{Z})\) of a receding galaxy and its distance \(r\) from earth are related as:
1 \(\mathrm{Z} \propto \mathrm{r}\)
2 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}}\)
3 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}^2}\)
4 \(\mathrm{Z} \propto \mathrm{r}^{3 / 2}\)
Explanation:
: According to Hubble's law, \(\mathrm{Z}=\mathrm{Hr}\) Where, \(\mathrm{H}=\) Hubble constant \(r=\) Distance from earth \(\therefore \quad \mathrm{Z} \propto \mathrm{r}\)
BCECE-2005
WAVE OPTICS
283087
A light of intensity \(I_0\) passes through a material of thickness \(d\), then the intensity will be
: We know that, Intensity \((\mathrm{I})=\mathrm{I}_0 \mathrm{e}^{-\mathrm{d} \lambda}\) Where, \(\mathrm{I}_0=\) Intensity of light \(\mathrm{d}=\) Thickness of material When a light passes through a material its intensity goes on decreasing. The above equation exponentially decreasing intensity. Whereas, option (b) and (c) represent increasing intensity.
AIIMS-2000
WAVE OPTICS
283093
The width of the diffraction band varies:
1 inversely as the wavelength
2 directly as the width of the slit
3 directly as the distance between the slit and the screen
4 inversely as the size of the source from which the slit is illuminated
Explanation:
: Width of lift pattern \((x)=\frac{(\mathrm{n} \lambda) \mathrm{D}}{\mathrm{a}}\) Where, \(\mathrm{D}=\) distance between screen and slit \(\mathrm{a}=\) width of slit \(\lambda=\) wavelength of light \(\mathrm{n}=\) order of diffraction So, varies directly as the distance between the slit and the screen.
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVE OPTICS
283083
In an electron microscope the accelerating voltage is increased from \(20 \mathrm{kV}\) to \(80 \mathrm{kV}\), the resolving power of the microscope will become
1 \(2 \mathrm{R}\)
2 \(\frac{R}{2}\)
3 \(4 \mathrm{R}\)
4 \(3 \mathrm{R}\)
Explanation:
: Wavelength and accelerating voltage for an electron, \(\lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{V}=\text { potential applied })\) \(\because\) Resolving power \(\propto \frac{1}{\lambda}\) \(\therefore\) Resolving power \(\propto \sqrt{\mathrm{V}}\) Now, if potential used is increased 4 times, resolving power will be increased 2 times. So, if resolving power earlier is R. then it becomes \(2 \mathrm{R}\).
AIIMS-2000
WAVE OPTICS
283085
According to Hubble's law, the red-shift \((\mathrm{Z})\) of a receding galaxy and its distance \(r\) from earth are related as:
1 \(\mathrm{Z} \propto \mathrm{r}\)
2 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}}\)
3 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}^2}\)
4 \(\mathrm{Z} \propto \mathrm{r}^{3 / 2}\)
Explanation:
: According to Hubble's law, \(\mathrm{Z}=\mathrm{Hr}\) Where, \(\mathrm{H}=\) Hubble constant \(r=\) Distance from earth \(\therefore \quad \mathrm{Z} \propto \mathrm{r}\)
BCECE-2005
WAVE OPTICS
283087
A light of intensity \(I_0\) passes through a material of thickness \(d\), then the intensity will be
: We know that, Intensity \((\mathrm{I})=\mathrm{I}_0 \mathrm{e}^{-\mathrm{d} \lambda}\) Where, \(\mathrm{I}_0=\) Intensity of light \(\mathrm{d}=\) Thickness of material When a light passes through a material its intensity goes on decreasing. The above equation exponentially decreasing intensity. Whereas, option (b) and (c) represent increasing intensity.
AIIMS-2000
WAVE OPTICS
283093
The width of the diffraction band varies:
1 inversely as the wavelength
2 directly as the width of the slit
3 directly as the distance between the slit and the screen
4 inversely as the size of the source from which the slit is illuminated
Explanation:
: Width of lift pattern \((x)=\frac{(\mathrm{n} \lambda) \mathrm{D}}{\mathrm{a}}\) Where, \(\mathrm{D}=\) distance between screen and slit \(\mathrm{a}=\) width of slit \(\lambda=\) wavelength of light \(\mathrm{n}=\) order of diffraction So, varies directly as the distance between the slit and the screen.
283083
In an electron microscope the accelerating voltage is increased from \(20 \mathrm{kV}\) to \(80 \mathrm{kV}\), the resolving power of the microscope will become
1 \(2 \mathrm{R}\)
2 \(\frac{R}{2}\)
3 \(4 \mathrm{R}\)
4 \(3 \mathrm{R}\)
Explanation:
: Wavelength and accelerating voltage for an electron, \(\lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{V}=\text { potential applied })\) \(\because\) Resolving power \(\propto \frac{1}{\lambda}\) \(\therefore\) Resolving power \(\propto \sqrt{\mathrm{V}}\) Now, if potential used is increased 4 times, resolving power will be increased 2 times. So, if resolving power earlier is R. then it becomes \(2 \mathrm{R}\).
AIIMS-2000
WAVE OPTICS
283085
According to Hubble's law, the red-shift \((\mathrm{Z})\) of a receding galaxy and its distance \(r\) from earth are related as:
1 \(\mathrm{Z} \propto \mathrm{r}\)
2 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}}\)
3 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}^2}\)
4 \(\mathrm{Z} \propto \mathrm{r}^{3 / 2}\)
Explanation:
: According to Hubble's law, \(\mathrm{Z}=\mathrm{Hr}\) Where, \(\mathrm{H}=\) Hubble constant \(r=\) Distance from earth \(\therefore \quad \mathrm{Z} \propto \mathrm{r}\)
BCECE-2005
WAVE OPTICS
283087
A light of intensity \(I_0\) passes through a material of thickness \(d\), then the intensity will be
: We know that, Intensity \((\mathrm{I})=\mathrm{I}_0 \mathrm{e}^{-\mathrm{d} \lambda}\) Where, \(\mathrm{I}_0=\) Intensity of light \(\mathrm{d}=\) Thickness of material When a light passes through a material its intensity goes on decreasing. The above equation exponentially decreasing intensity. Whereas, option (b) and (c) represent increasing intensity.
AIIMS-2000
WAVE OPTICS
283093
The width of the diffraction band varies:
1 inversely as the wavelength
2 directly as the width of the slit
3 directly as the distance between the slit and the screen
4 inversely as the size of the source from which the slit is illuminated
Explanation:
: Width of lift pattern \((x)=\frac{(\mathrm{n} \lambda) \mathrm{D}}{\mathrm{a}}\) Where, \(\mathrm{D}=\) distance between screen and slit \(\mathrm{a}=\) width of slit \(\lambda=\) wavelength of light \(\mathrm{n}=\) order of diffraction So, varies directly as the distance between the slit and the screen.
283083
In an electron microscope the accelerating voltage is increased from \(20 \mathrm{kV}\) to \(80 \mathrm{kV}\), the resolving power of the microscope will become
1 \(2 \mathrm{R}\)
2 \(\frac{R}{2}\)
3 \(4 \mathrm{R}\)
4 \(3 \mathrm{R}\)
Explanation:
: Wavelength and accelerating voltage for an electron, \(\lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \quad(\because \mathrm{V}=\text { potential applied })\) \(\because\) Resolving power \(\propto \frac{1}{\lambda}\) \(\therefore\) Resolving power \(\propto \sqrt{\mathrm{V}}\) Now, if potential used is increased 4 times, resolving power will be increased 2 times. So, if resolving power earlier is R. then it becomes \(2 \mathrm{R}\).
AIIMS-2000
WAVE OPTICS
283085
According to Hubble's law, the red-shift \((\mathrm{Z})\) of a receding galaxy and its distance \(r\) from earth are related as:
1 \(\mathrm{Z} \propto \mathrm{r}\)
2 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}}\)
3 \(\mathrm{Z} \propto \frac{1}{\mathrm{r}^2}\)
4 \(\mathrm{Z} \propto \mathrm{r}^{3 / 2}\)
Explanation:
: According to Hubble's law, \(\mathrm{Z}=\mathrm{Hr}\) Where, \(\mathrm{H}=\) Hubble constant \(r=\) Distance from earth \(\therefore \quad \mathrm{Z} \propto \mathrm{r}\)
BCECE-2005
WAVE OPTICS
283087
A light of intensity \(I_0\) passes through a material of thickness \(d\), then the intensity will be
: We know that, Intensity \((\mathrm{I})=\mathrm{I}_0 \mathrm{e}^{-\mathrm{d} \lambda}\) Where, \(\mathrm{I}_0=\) Intensity of light \(\mathrm{d}=\) Thickness of material When a light passes through a material its intensity goes on decreasing. The above equation exponentially decreasing intensity. Whereas, option (b) and (c) represent increasing intensity.
AIIMS-2000
WAVE OPTICS
283093
The width of the diffraction band varies:
1 inversely as the wavelength
2 directly as the width of the slit
3 directly as the distance between the slit and the screen
4 inversely as the size of the source from which the slit is illuminated
Explanation:
: Width of lift pattern \((x)=\frac{(\mathrm{n} \lambda) \mathrm{D}}{\mathrm{a}}\) Where, \(\mathrm{D}=\) distance between screen and slit \(\mathrm{a}=\) width of slit \(\lambda=\) wavelength of light \(\mathrm{n}=\) order of diffraction So, varies directly as the distance between the slit and the screen.