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MCQs on Diffraction of light(set-004)
Quiz
- Two spectral lines, incident on a grating of grating element $a+b$ have wavelengths $\lambda$ and $\lambda+ d \lambda$ respectively.For normal incidence, the angular separation between these lines in $n^{th}$ order must be
- $\frac{n d \lambda}{(a+b)- n \lambda}$
- $\frac{n d \lambda}{\sqrt{(a+b)^2+ n^2 \lambda^2}}$
- $\frac{n d \lambda}{\sqrt{(a+b)^2- n^2 \lambda^2}}$
- $\frac{n d \lambda}{(a+b)^2- n^2 \lambda^2}$
- A parallel beam of monochromatic light of wavelength $4000A^o$ is incident normally on a grating of grating element length $15000A^o$. The angular dispersion of the diffracted beam in the third order spectrum must be
- $3.34 \times 10^{-1} rad/A^o$
- $3.34 \times 10^{-2} rad/A^o$
- $3.34 \times 10^{-3} rad/A^o$
- $3.34 \times 10^{-4} rad/A^o$
- In the fraunhoffer diffraction pattern formed by a single slit, the slit width is $0.04 cm$ and wawveleingth of light used is $5000 A^o$. The diffraction angle for the first dark fringe must be
- $1.25 \times 10^{-3} rad$
- $2.25 \times 10^{-3} rad$
- $3.25 \times 10^{-3} rad$
- $4.25 \times 10^{-3} rad$
- A convex lens of focal length 25 cm is placed after a slit of width 0.5 mm . If a plane wave of wavelength $5000 A^o$ falls on the slit normally, then, the separation between the second minima on either side of the central maximum of the diffraction pattern,would be
- $1.0 \times 10^{-3}$
- $1.5 \times 10^{-3}$
- $2.0 \times 10^{-3}$
- $2.5 \times 10^{-3}$
- Fringes are seen within the shadow of a thin wire held vertically in front of a narrow slit illumnated by light of wavelength $5000 A^o$. If the fringe width at a distance of 3 m is 1.5 mm, the diameter of the wire has to be
- 0.5 mm
- 1 mm
- 1.5 mm
- 2 mm
- When a circular aperature of 1.5 mm radius is illuminated by a plane wave of light, the most intense point on the axis is at a distance of 350 cm from the aperature. The wavelength of the incident light must be
- $4428.5 A^o$
- $5428.5 A^o$
- $6428.5 A^o$
- $7428.5 A^o$
- A zone plate of focal length 21 cm is illuminated by light of wavelength $4200 A^o$. The radius of the $11^{th}$ zone of the zone plate must be
- $58.48 \times 10^{-5} m $
- $78.48 \times 10^{-5} m $
- $98.48 \times 10^{-5} m $
- $118.48 \times 10^{-5} m $
- A grating having 4500 lines per cm is illuminated by a light of wavelength $4000 A^o$. The highest order of the spectrum , that will be visible , will be
- 5
- 6
- 3
- 4
- The pattern of netwton rings formed between a plane glass plate and a convex lens of focal length180 cm , is used as a zone plate. The principal focal length of the zone plate will be
- 2.8 m
- 1.8 m
- 4.8 m
- 3.8 m
- The image of a point source of light of wavelength $4800 A^o$ at a distance of 1.2 m from the zone plate is observed at 2.0 m on the other side. The focal lenth of the zone plate must be
- 85 cm
- 75 cm
- 65 cm
- 55 cm
MCQs on Dynamics of Charged Particles(Set-003)
Dynamics of Charged Particles
Quiz
- The intensity of the electric field which will impart an acceleration equal to acceleration due to gravity to an alpha particle initially at rest
- $20.46 \times 10^{-7} N/C $
- $20.46 \times 10^{-8} N/C $
- $20.46 \times 10^{-6} N/C $
- $20.46 \times 10^{-9} N/C $
- If a beam of charged particles moving with initiaol velocity of $7.5 \times 10^6 m/s$ , gets deflected by $4 mm$ in traversing a distance of 12 cm througfh an electrostatic field of $2.2 kV/m$ , perpendicular to their path, the specific charge of the particles would be
- $227.2 \times 10^6 C/kg$
- $227.2 \times 10^3 C/kg$
- $227.2 C/kg$
- None above
- A helium nucleus was initially at rest. After being accelerated by a PD of 2.8kV,it passes through a region $AB$ of length 5 m with uniform velocity. The transit time required by the nucleus to traverse the length $AB$, must be
- $ 1.93 \mu s$
- $ 1.93 m s$
- $ 1.93 n s $
- $ 1.93 p s$
- Two parallel plates are kept in air at 2 cm apart and a PD of 180 V is applied.An electron was resting initially on the plate at lower potential. The KE of the electron when it will reach the other plate, have to be
- $1.79 \times 10^2 eV$
- $1.79 \times 10^3 eV$
- $1.79 \times 10^4eV$
- $1.79 \times 10^5 eV$
- A proton travelling with a velocity of $v_x= 6 \times 10^8 m /s $ enters in to a region of uniform magnetic field $\vec{B}= 3000 \hat{j}-1800 \hat{k}$ Gauss. It will experience a force of
- $(28.8 \hat{i}+17.28 \hat{j}) \times 10^{-8} N$
- $(28.8 \hat{i}-17.28 \hat{j}) \times 10^{-8} N$
- $(28.8 \hat{k}-17.28 \hat{j}) \times 10^{-8} N$
- $(28.8 \hat{k}+17.28 \hat{j}) \times 10^{-8} N$
- An electron of velocity $v= (3 \hat{j}- 2 \hat{k}) \times 10^6 \ cm/s$ enters a region of uniform magnetic field , $B= 450 \hat{j} Gauss$. Its path inside the magnetic field would be
- elliptical path
- Circular path
- helical path
- Straight line inclined at an angle to the direction of the magnetic field
- An electron has a velocity of $7.5 \times 10^5 m/s$ normal to a magnetic field of $2.5 Wm/m^2$. The frequency of revolution of the electron inside the magentic field, must be
- $7.0004 \times 10^{7} \ \ rps$
- $7.0004 \times 10^{8} \ \ rps$
- $7.0004 \times 10^{19} \ \ rps$
- $7.0004 \times 10^{10} \ \ rps$
- A proton, after being projected towards a conductor at positive potential, can manage just only to strike the conductor at $+1500 V $ potential.The initial velocity of the proton must be
- $54.77 \times 10^3 m/s$
- $54.77 \times 10^4 m/s$
- $54.77 \times 10^5 m/s$
- $54.77 \times 10^6 m/s$
- An electron of energy $25 eV$ moving in a direction perpendicular to a magnetic field of $B=120 Wb/m^2$, describes a circular path . The radius of the circular path would be
- $0.140\times 10^{-5} m $
- $0.140\times 10^{-6} m $
- $0.140\times 10^{-7} m $
- $0.140\times 10^{-8} m $
- The electric field between the plates of a CRO is $1.8 \times 10^4 N/C$. The plates are 2 cm long.The deflection that an electron will experience if it enters at right angles to the field with a KE of 2.5KeV, will be
- 18 cm
- 1.8 cm
- 0.18 cm
- 0.018 cm
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