Question

Ascreen is placed 40 cm away from an illuminated object. a a converging lens is placed between the source and the screen and and an attempt is made to form the image of the source on the screen . if you know position of the lens could be found. the focal length of the lens is

a) must be less than 10 cm
b) must must be greater than 10 cm
c) must not be greater than 20 cm
d) must not be less than 10 cm

​

Answer 1

Ascreen is placed 40 cm away from an illuminated object. a a converging lens is placed between the source and the screen and and an attempt is made to form the image of the source on the screen . if you know position of the lens could be found. the focal length of the lens is

Ascreen is placed 40 cm away from an illuminated object. a a converging lens is placed between the source and the screen and and an attempt is made to form the image of the source on the screen . if you know position of the lens could be found. the focal length of the lens isa) must be less than 10 cm✔️

b) must must be greater than 10 cm

b) must must be greater than 10 cmc) must not be greater than 20 cm

b) must must be greater than 10 cmc) must not be greater than 20 cmd) must not be less than 10 cm

Answer 2

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

Let, an image is formed in the screen.

Let, object distance = u

therefore, Image distance, v = 40 – u

Now, from lens formula

we have,

1/v – 1/u = 1/f

=> 1/(40 – u) – 1/(-u) = 1/fmin

=> 1/(40-u) + 1/u = 1/fmin ….....(i)

Differentiating with respect to ‘u’ ,

=> d/du{1/(40-u)}+ d/du {1/u) = d/du{1/fmin}

=>(40-u)-2 - u-2 = – f-2 d/du {fmin}

=> (40-u)-2- u-2 = 0 [since d/du {fmin} = 0]

=>(40-u)-2= u-2

=> 1/(1600 – 80u + u2) = 1/u2

=>80u = 1600

=> u = 20

(i)=> 1/20 + 1/(40-20) = 1/fmin

=>fmin = 10

Therefore, the minimum focal length is 10 when the image is form.

But according to question the image can’t be form, so the focal length of the image must be greater than 10 .