Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be :

Find the ratio of maximum intensity to the minimum intensity in the interference pattern if the widths of the two slits in Young's experiment are in the ratio of 9 : 16. (Assuming intensity of light is directly proportional to the width of slits)

In Young's double slit experiment, the fringe width is $$12\mathrm{mm}$$. If the entire arrangement is placed in water of refractive index $$\frac{4}{3}$$, then the fringe width becomes (in mm):

Two coherent sources of light interfere. The intensity ratio of two sources is $$1:4$$. For this interference pattern if the value of $$\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }}$$ is equal to $$\frac{2\alpha +1}{\beta +3}$$, then $$\frac{\alpha }{\beta }$$ will be :

An unpolarised light beam of intensity $$2I_0$$ is passed through a polaroid P and then through another polaroid Q which is oriented in such a way that its passing axis makes an angle of $$30^{\circ }$$ relative to that of P. The intensity of the emergent light is

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

Two capacitors having capacitance C1 and C2 respectively are connected as shown in figure. Initially, capacitor C1 is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C1 is now connected to uncharged capacitor C2 by closing the switch S. The amount of charge on the capacitor C2, after equilibrium, is :

The charge on capacitor of capacitance 15$$\mu$$F in the figure given below is :

A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 $$\mu$$F. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be :

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness {d \over 2} and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

A parallel plate capacitor is formed by two plates each of area 30$$\pi$$ cm2 separated by 1 mm. A material of dielectric strength 3.6 $$\times$$ 107 Vm$$-$$1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 $$\times$$ 10$$-$$6C, the value of dielectric constant of the material is : [Use {1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9} Nm2 C$$-$$2]

Co is the capacitance of a parallel plate capacitor with air as a medium between the plates (as shown in Fig. 1). If half space between the plates is filled with a dielectric of relative permittivity $$\epsilon$$r (as shown in Fig. 2), the new capacitance of the capacitor will be :

A condenser of $$2\mu \mathrm{F}$$ capacitance is charged steadily from 0 to $$5\mathrm{C}$$. Which of the following graph represents correctly the variation of potential difference $$(\mathrm{V})$$ across it's plates with respect to the charge $$(Q)$$ on the condenser?

Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii \left( {{{{R_2}} \over {{R_1}}}} \right) is :

The total charge on the system of capacitors $$C_1=1\mu \mathrm{F},C_2=2\mu \mathrm{F},{\mathrm{C}}_3=4\mu \mathrm{F}$$ and $${\mathrm{C}}_4=3\mu \mathrm{F}$$ connected in parallel is : (Assume a battery of $$20\mathrm{V}$$ is connected to the combination)

A source of potential difference $$V$$ is connected to the combination of two identical capacitors as shown in the figure. When key '$$K$$' is closed, the total energy stored across the combination is $$E_1$$. Now key '$$K$$' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $$E_2$$. The ratio $$E_1/E_2$$ will be :

Two capacitors, each having capacitance $$40\mu \mathrm{F}$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$\mathrm{K}$$ such that the equivalence capacitance of the system became $$24\mu \mathrm{F}$$. The value of $$\mathrm{K}$$ will be :