Question

 A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.​

Answer 1

Answer:

A concave lens has a focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Explanation:

f = -20cm

h1 = 5 cm

v = -15 cm (Concave lens forms virtual image)

1/v – 1/u = 1/f

1/-15 – 1/u = 1/-20

1/u = -(1/15) + 1/20

1/u = -1/60

u = -60 cm

object should be placed 60 cm to the left of the lens.

m = v/u = h2/h1

-15/-60 = h2/5

h2 = 1.25 cm

Answer 2

☆GIVEN:

¤ A concave lens whose;

focal length = - 20 cm (always negative )

object size = 5 cm

Image distance = -15 cm ( as image formed is always virtual and on the same side of the object )

☆TO FIND :

1. object distance

2. Image size

☆SOLUTION :

We use the lens formula;

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

where;

v = image distance

u = object distance

f = focal length

putting the values in the formula;

$( \frac{ - 1}{15} - \frac{1}{u})cm = \frac{ - 1}{20} cm$

$- \frac{1 }{u} = \frac{ - 1}{20}cm + \frac{1}{15} cm \\ \\ \frac{ - 1}{u} = \frac{ - 15 + 20}{15 \times 20} cm \\ \\ \frac{ - 1}{u} = \frac{5}{15 \times 20} cm \\ \\ \frac{ - 1}{u} = \frac{1}{3 \times 20} cm \\ \\ \frac{ - 1}{ u } = \frac{1}{60} cm \\ \\ u = - 60cm$

$object \: distance = - 60cm$

(negative sign indicates that it is formed on the left side of the pole)

now,

$\frac{height \: of \: image}{height \: of \: object} = \frac{image \: distance}{objet \: distance}$

putting the value in the formula;

(we put the values with sign)

$\frac{h(i)}{5cm} = \frac{ - 15cm}{ - 60cm} \\ \frac{h(i)}{5cm} = \frac{1}{4} \\ \\ h(i) = \frac{5}{4} cm$

$height \: of \: image = \frac{5}{4} cm$

☆Additional information:

Nature of the image in this case is;

1. Virtual and erect

2. enlarged