Question

(a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of
(i) 0.50 m
(ii) 0.25 m
(iii) 0.15 m
(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?

Answer 1

Explanation:

(i)  Step 1:

Given

u = -0.50 m = -50 cm  

= 0.20 = 20  

 $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$                         by lens formula

$\frac{1}{20} = \frac{1}{v} - \frac{1}{-50}$

$\frac{1}{v} =\frac{1}{20}- \frac{1}{50}$

$\frac{1}{v} = \frac{5-2}{100}$

$v = \frac{100}{3}$ cm

The image on the right side of the mirror magnification will be at a distance of 33.3 cm

Step 2:

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

$m= \frac{33.3}{-50}$

m= -0.66

Magnification is negative so image is real and inverted.

(ii) Step 1:

Given

u = -25 cm  

f = 20 cm  

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$                    by lens formula

$\frac{1}{20} = \frac{1}{v} - \frac{1}{-25}$

$\frac{1}{v} = \frac{5-4}{100}$

$\frac{1}{v} = \frac{1}{100}$

v=100 cm  

The picture is shaped 100 cm behind the mirror.

Step 2:

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

$m = \frac{100}{(-25)}$

m = -4

Magnification is negative, meaning that the image is real and is inverted.

(c) Step 1:

Given u = -15 cm  

     f = 20  

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$              by lens formula

$\frac{1}{20} =\frac{1}{v} - \frac{1}{(-15)}$

$\frac{1}{v} = (\frac{(3-4)}{60} )$

v = -60 cm

The picture will be shaped in front of the mirror 60 cm away.

Step 2:

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

m= $\frac{-60}{-15}$

m =  4

Magnification is good, so the image is both virtual and erect.

(b)  For a film projector we can use case (ii)

    For a camera, we may use case (i)

    For a magnifying glass we may use case (ii).

You can get all the answers of Chapter 5 in the link below:    

https://brainly.in/question/14963222

   

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