Category Archives: Worked Examples

Lens Equation

A converging lens is used to magnify a virtual image to four times it original size. Given then focal length of the converging lens is 20.0 cm, what is the distance of the object from the lens?

m=v/u

4=v/u   =>  v=4u

But image is virtual, so v must be negative (ie. v = -4u)

1/f = 1/u +1/v

1/f = 1/u + 1/(-4u)

1/20 = (4-1)/4u

u = 15 cm

In this question it is important that you recognise the image is a virtual image and thus v has to be a negative number.

 

Charge moving through magnetic and electric fields

The magnitude of the force created by the electric field is given by

FE = Eq

FE = EQ (using the values in this question)

The magnitude of the force created by the magnetic field is given by

FM = Bqv

FM = BQv (using the values in this question)

 

Q could be +ve or -ve.

If Q is +ve, then:

    • the electric field will exert a downward force on Q
    • the magnetic field will exert an upwards force on Q (from Fleming’s Left Hand Rule)

This could still create the motion indicated in the diagram, however FM would need to be smaller than FE.

FM < FE

BQv < EQ

v < E/Q

If Q is –ve, then:

    • the electric field will exert an upwards force on Q
    • the magnetic field will exert a downwards force on Q (from Fleming’s Left Hand Rule)

This could still create the motion indicated in the diagram, however FM would need to be larger than FE.

FM > FE

BQv > EQ

v > E/Q

Which is answer D