Question

An object placed of a distance of 30 cm from a concave lens of focal length 30cm find as land formula and find the distance of image from the lens determine natura position size draw labelled diagram

Answer 1

_________________________________

$\underline{\boxed{\frak{solution}}}$

Concave lens

position of object, u = -30 cm

focal length of concave lens, f = -30 cm

let,

position of image, v = ?

using lens formula

$\bf \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ \bf \: \frac{1}{ - 30} = \frac{1}{v} - \frac{1}{ - 30} \\ \\ \bf \: \frac{1}{ - 30} = \frac{1}{v} + \frac{1}{30} \\ \\ \bf \frac{1}{v} = \frac{1}{ - 30} - \frac{1}{30} \\ \\ \bf \: \frac{1}{v} = \frac{ - 1 - 1}{30} = \frac{ - 2}{30} \\ \\ \bf \: v = - 15 \: cm$

So, distance of image from lens will be 15 cm in the negative axis.

now,

magnification, m = v / u

m = -15 / -30

m = 0.5

Hence the image formed will be virtual and erect as well as smaller than object in size.

_________________________________

Answer 2

Answer:

hello friends

Explanation:

Concave lens

position of object, u = -30 cm

focal length of concave lens, f = -30 cm

let,

position of image, v = ?

using lens formula

f

1

=

v

1

−

u

1

−30

1

=

v

1

−

−30

1

−30

1

=

v

1

+

30

1

v

1

=

−30

1

−

30

1

v

1

=

30

−1−1

=

30

−2

v=−15cm

So, distance of image from lens will be 15 cm in the negative axis.

now,

magnification, m = v / u

m = -15 / -30

m = 0.5

Hence the image formed will be virtual and erect as well as smaller than object in size.

_________________________________