Question

An object of height 7.5 cm is placed at a distance of 35 cm from the
converging lens whose focal length is 15 cm. (i) what is the height of the
image formed? (ii) what is the magnification produced by the lens? (iii)
what is the nature of the image formed by the lens?

Answer 1

GIVEN THAT -

Height of object ( O ) = 7.5 cm

distance of object ( U ) = - 35 cm

Focal length ( F ) = + 15 cm

Distance of Image ( V ) = ?

Height of Images ( I ) = ?

Magnification ( M ) = ?

By lens formula ,

1 / F = 1 / V - 1 / U

1 / 15 = 1 / V - ( 1 / - 35 )

1 / 15 = 1 / V + 1 / 35

1 / V = 1 / 15 - 1 / 35

1 / V = ( 7 - 3 ) / 105

1 / V = 4 / 105

V = 105 / 4 = 26.25 cm

M = I / O = V / U

I / 7.5 = 26.25 / -35

I = - 0.75 x 7.5 = -5.625 cm

M = I / O = - 5.625 / 7.5 = - 0.75

Image is real inverted and diminished....

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Answer 2

Answer :

• height of the image formed = -5.6 cm
• magnification produced by the lens = -0.75
• The image formed is real and inverted and diminished.

Explanation. :

Given :

Height of the object ( O) = 7.5 cm

Object distance (u) = -35 cm

Focal length (f) = 15 cm

image distance (v) = ?

image distance ( i) = ?

Formula to be Used :

Lens formula :

$\frac{1}{f } = \frac{1}{v} - \frac{1}{u \: } \\ \frac{1}{15 } = \frac{1}{v} - \frac{1}{( - 35)} \\ \frac{1}{v} = \frac{1}{15} - \frac{1}{35} \\ \frac{1}{v} = \frac{4}{105} \\ v = \frac{105}{4} = 26.25cm$

Now, Calculating magnification:

$m \: = \frac{i}{o} = \frac{v}{u} \\ \frac{i}{7.5} = \frac{26.25}{( - 35)}$

$i \: = - 5.625$

$m \: = \frac{i}{o} = \frac{ - 5.625}{7.5} = - 0.75$

Hence,

(i) height of the image formed = -5.6 cm

(ii) magnification produced by the lens = -0.75

(iii) The image formed is real and inverted since it's magnification is negative and it is diminished because the magnification is less than one.

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