Question

A 5-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image​

Answer 1

Explanation:

Here,

O=5cm

u=5cm (negative according to the cartesian rule)

f=15cm (for concave mirror)

By relation,

1/f=1/u+1/v

=>1/-15=1/-5+1/v

=>1/-15=(v-5)/-5v

=>-5v=-15v+75

=>10v=75

=>v=7.5

Therefore the image is formed 7.5 cm behind the mirror i.e; a virtual and erect image is formed.

Magnification (M) =-(v/u)

=>M=-(7.5/-5)

=> M = 1.5

Therefore the image is magnified 1.5 times.(ans)

Also,

M=I/O

=> 1.5=I/5

=> I = 7.5 cm

Therefore , image height = 7.5 cm

Answer 2

Correct Question :-

A 5-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image

$\sf Given: - \begin{gathered} \\ \end{gathered}$

The focal length (f) = 15 cm

The plus sign indicates that the focal point is real or the rays pas through the point.

The object height (ho) = 5 cm

The object distance (do) = 5 cm

Wanted :-

The image distance, the magnification of image, the image height and the properties of image

Explanation :-

The image distance (di) :

1/di = 1/f – 1/do = 1/15 – 1/5 = 1/15 – 3/15 = -2/15

di = -15/2 = -7.5 cm

The minus sign indicates that the image is virtual or the rays do not pas through the image.

The magnification of image (m) :

m = – di / do = -(-7.5)/5 = 7.5/5 = 1.5

The plus sign indicates that the image is upright.

The image height (hi) :

m = hi / ho

hi = m ho = (1.5)5 = 10/3 = 7.5 cm

The plus sign indicates that the image is upright.

The properties of the image :-

● virtual

● upright

● The image greater than the object

● The image distance is greater than the object distance

Hope it help you