Question

(l) Two lenses have power of
(a) +2d. (b) -4d

What is the nature and focal length of each lense?

(ll) An object is kept at a distance of 100 cm from each of the above lenses. Calculate
(a) image distance
(b) magnification in each of the two cases.

Answer 1

HELLO DEAR,

we know that:-

f=1/p

where,

p=power

f = focal length

(1) :-

(a) given that;-

p=+2D

using the the formula

we get,

1/f=2

f=1/2

f=0.5m =50cm


the lens is convex lens of focal length= 0.5m = 50cm


(b) given that;-

p= -4D

using the the formula

we get,

1/f= -4

f= 1/-4

f= -0.25m = -25cm

the lens is concave lens of focal length= -0.25m = -25cm



(2) given that;-

u = -100cm = -1 m
and f=( 0.5m) for first lens

we know that:-

lens Formula

$= > \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
now put the values

we get,

$\frac{1}{0.5} = \frac{1}{v} - \frac{1}{ - 1} \\ = > \frac{1}{v} = \frac{1}{0.5} - \frac{1}{1} \\ = > \frac{1}{v} = \frac{1 \times 10}{0.5 \times 10} - 1 \\ = > \frac{1}{v} = \frac{10}{5} - 1 \\ = > \frac{1}{v} = 2 - 1 = 1 \\ = > v = 1$
and

magnification=

$m = \frac{v}{u} \\ = > m= \frac{1}{ - 1} = - 1$


(b) for second lens
given that;-

u = -100cm = -1 m

and f= (-0.25m) for second lens

we know that:-

lens Formula


$= > \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{ - 0.25} = \frac{1}{v} - \frac{1}{ - 1} \\ = > \frac{1}{v} = \frac{1}{ - 0.25} - 1 \\ = > \frac{1}{v} = - \frac{1 \times 100}{ 0.25 \times 100} - 1 \\ = > \frac{1}{v} = - \frac{100}{25} - 1 \\ = > \frac{1}{v} = - 4 - 1 = - 5 \\ = > v = - \frac{1}{5} = - 0.2$

and

magnification is

$m = \frac{v}{u} = \frac{ - 0.2}{ - 1} \\ = > m = + 0.2$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

"Magnification, M = 0.2

Solution:

For +2D lens,

we know that power $=\frac { 1 }{ focal\quad length } \quad in\quad \left( m \right)$

$+2D = \frac { 1 }{ f }$

$F = \frac { 1 }{ 2 } D$

F = 0.5m = 50cm (it's a convex lens)

U =  -100cm (using sign conventions)

Using lens formula,

$\frac { 1 }{ f } \quad =\quad \frac { 1 }{ v } -\frac { 1 }{ u }$

$\frac { 1 }{ 50 } \quad =\quad \frac { 1 }{ v } -\frac { 1 }{ -100 }$

$\frac { 1 }{ 50 } \quad =\quad \frac { 1 }{ v } +\frac { 1 }{ 100 }$

$\frac { 1 }{ 50 } -\frac { 1 }{ 100 } \quad =\quad \frac { 1 }{ v }$

$\frac { 1 }{ 100 } =\frac { 1 }{ v }$

v=100cm

since v is +ve, the image is real.

Magnification ${ M }=\frac { v }{ u }$

${ M }=\frac { 100 }{ -100 } \quad \quad \quad \Rightarrow M = -1$

For -4D lens,

we know that $=\frac { 1 }{ focal\quad length } \quad in\quad \left( m \right) -4{ D }=\frac { 1 }{ f }$

${ f }=-\frac { 1 }{ 4D }$

F = -0.25m = -25cm (it's a concave lens)

U = -100cm (using sign conventions)

Using lens formula,

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

$\frac { 1 }{ -25 } =\frac { 1 }{ v } -\frac { 1 }{ -100 }$

$\frac { 1 }{ -25 } =\frac { 1 }{ v } +\frac { 1 }{ 100 }$

$\frac { 1 }{ -25 } -\frac { 1 }{ 100 } =\frac { 1 }{ v }$

$-4-\frac { 1 }{ 100 } =\frac { 1 }{ v }$

$\frac { -5 }{ 100 } =\frac { 1 }{ v }$

$\frac { 1 }{ v } =\frac { -1 }{ 20 }$

v=-20cm

since v is -ve, image is virtual.

Magnification ${ M }=\frac { v }{ u }$

${ M }=\frac { -20 }{ -100 }$

M = 0.2"