Question

(a) Two lenses have power of (i) + 2D (ii) - 4D. What is the nature and focal length

of each lens.?

(b) An object is kept at a distance of 100cm from each of the above lenses.

Calculate the (i) image distance (ii) magnification in each of the two cases.

Answer 1

(A)
$\huge {\bold {\red{Nature\:and \:Focal \:length}}}$

f=1/p

where,

p=power

f = focal length

(1)
given

p= +2

f = 1/2=.5m

value is positive so lens is convex with focal length 0.5m

(2)

given
p=-4

f= 1/-4= -0.25m

value is negative so lens is concave of focal length -0.25m.

(B
$\huge {\bold{\red{Image\ Distance}}}$

(1)

We know that

=>> $\huge \frac{1}{f} =\frac{1}{v} -\frac{1}{u}$


given

u= -100cm = -1m
f= 0.5m

1/v= 1/f +1/u

1/v= 1/0.5+ 1/-1

1/v= 2-1

1/v=1

v=1

Image distance =1m

(2)

u= -100=-1m
f= -0.25

1/v=1/f +1/u
1/v = 1/-0.25 + 1/-1

1/v = -5

v= 1/-5=-0.2

Image distance = -0.2m

(C)
$\huge {\bold{\red {Magnification}}}$

(1st)

We know that

Magnification

m= v/u
m=1/-1
m =-1

(2nd)

m=v/u
m= -0.2/-1
m= .02