Question

A 5cm high object is placed at a distance of 50cm from a concave lens, having a focal length of 25cm. Find the height and position of the image so formed.

Answer 1

SOLUTION:-

Given:

Distance of object from lens

u= -50cm

f= -25cm

h= 5cm

Therefore,

Using Formula by,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f '}, \: we \: get \\ \\ = > \frac{1}{v} + \frac{1}{50} = \frac{1}{ - 25 } \\ \\ = > \frac{1}{v} = - \frac{1}{25} - \frac{1}{50} \\ \\ = > \frac{1}{v} = - \frac{3}{50} \\ \\ = > 3v = - 50 \\ \\ = > v = - \frac{50}{3} \\ \\ = > v = 16 .67cm$

So, image will be formed on the same side of object at a distance of 16.67cm from the lens.

Again,

$m = \frac{h'}{h} = \frac{v}{u} \\ \\ = > I= \frac{v}{u} \times h$

Therefore,

Height of image, h'.

$= > \frac{ \frac{ - 50}{3} }{ - 50} \times 5 \\ \\ = > \frac{5}{3} \\ \\ = > 1.67cm \: \: \: (erect)$

Hope it helps ☺️

Answer 2

Answer:

3.3 cm, Real) Solve the example

Ask for details Follow Report by Adriano2915 26.06.2018

Answers

tiwaavi

tiwaavi Genius

Given conditions ⇒

Height of the Object(H₀) = 5 cm.

Object Distance(u) = 25 cm.

Focal length(f) = 10 cm.

Since, the lens is converging,

∴ u = -25 cm.

f = 10 cm.

Using the lens formula,

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

⇒ \frac{1}{10} = \frac{1}{v} - \frac{1}{-25}

⇒ v = 50/3

∴ v = 16.67 cm.

Since, the image distance is positive. This means that the real image is formed by the Lens. Also, It is formed on the Opposite size of the lens. I means on the side where the x-axis is present.

Now, Magnification = v/u

= -16.67/25

= - 0.6668

Also, Magnification = H₁/H₀

∴ H₁ = 0.6668 × 5

= 3.334 cm.

≈ 3.3 cm.

Since, the height of the image is less than than that of the object, therefore the image is diminished with respect to the object.

Hope it helps.