In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 $$\stackrel{o}{A}$$ is :-

With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 $$\stackrel{o}{A}$$ is observed at 5896 $$\stackrel{o}{A}$$ ?

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d0 + a0 sin$$\omega$$t; where d0, $$\omega$$ and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

In Young's double slit experiment, if the source of light changes from orange to blue then :

The light waves from two coherent sources have same intensity I1 = I2 = I0. In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima?

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :

An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '$$\alpha$$' with the plates. It leaves the plates at angle '$$\beta$$' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :

Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is {{15} \over 4} times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, {{{C_2}} \over {{C_1}}}.

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is {3 \over 4}d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as : $$\epsilon (x)={\epsilon }_0+kx$$, for \left( {0 < x \le {d \over 2}} \right) $$\epsilon (x)={\epsilon }_0+k(d-x)$$, for \left( {{d \over 2} \le x \le d} \right)

If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :

A capacitor of capacitance C = 1 $$\mu$$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $$\Omega$$. The time taken for the capacitor to be charged to get 50 V is : [Take ln 2 = 0.69]

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be : (Given area of plate = A)

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

Match List I with List II. List - I List - II (a) Capacitance, C (i) $$M^1{L}^1{T}^{-3}{A}^{-1}$$ (b) Permittivity of free space, $${\epsilon }_0$$ (ii) $$M^{-1}{L}^{-3}{T}^4{A}^2$$ (c) Permeability of free space, $${\mu }_0$$ (iii) $$M^{-1}{L}^{-2}{T}^4{A}^2$$ (d) Electric field, E (iv) $$M^1{L}^1{T}^{-2}{A}^{-2}$$ Choose the correct answer from the options given below

A simple pendulum of mass 'm', length 'l' and charge '+ q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

The material filled between the plates of a parallel plate capacitor has resistivity 200 $$\Omega$$m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50)

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

Calculate the amount of charge on capacitor of 4$$\mu$$F. The internal resistance of battery is 1$$\Omega$$ :

Three capacitors C1 = 2$$\mu$$F, C2 = 6$$\mu$$F and C3 = 12$$\mu$$F are connected as shown in figure. Find the ratio of the charges on capacitors C1, C2 and C3 respectively :