Question

a needle placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of the lens . identify the type of lens .determine its focal length and power . what is the size of the image if , needle is 5 cm in height?

Answer 1

hey that's the solution u require

Answer 2

Answer:

• It is a convex lens with a focal length of 30 cm and power of 0.033D.
• The size of the image is -10 cm when the needle is 5cm in height.

Explanation:

Given:

Object distance, u=−45cm,

Image distance, v=90cm,

Height of the object, O=5cm

Using mirror formula:

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

$\frac{1}{f} = \frac{1}{90} -\frac{1}{-45}$

$\frac{1}{f} = \frac{1+2}{90}$

$\frac{1}{f} = \frac{3}{90}$

$\frac{1}{f} = \frac{1}{30}$

$f= 30$

∴ Focal length = 30 cm

Now Power of the lens,

$P= \frac{1}{f}$

$P= \frac{1}{30}$

$P= 0.033 D$

∴The power of the lens is 0.033 D

Also magnification,

m = image distance/ object distance = height of image/ height of object

$m=\frac{v}{u} = \frac{I}{O}$

$I=\frac{90}{-45}$ ×$5$

$I= -10$

∴ Size of image = -10 cm

∴ it is a convex lens and the image is inverted.

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