An object is placed 120cm from a screen.. A lens of focal length 25 cm is used t

Question

An object is placed 120cm from a screen.. A lens of focal length 25 cm is used to produce an imagine the screen. (a) Find two locations for the lens which will magnification of the image on the screen.(b) For each of these locations calculate the magnification of the image, and show whether the image is real or virtual, erect or inverted .(c) For a lens of focal length 10cm , what is the minimum distance between object and screen that will allow an image to be produced?

QUESTION IS COMPLETE, IT IS WHAT THE INSTRUCTOR PROVIDED, I HAVE PART OF PART A CORRECT BUT NOT THE OTHER PARTS. HE MARKED PARTS B-C WRONG BECAUSE I DID NOT KNOW HOW TO DO IT.

2) An object is pl produce an jmage on the screen. () Find two locations for the lens which will produce animage on the screen. (b) For each of these locations cattate the magnification of thé image, and show whether the image is real or virtual, erect or aged 120 a screen. A lens of focal length 25 cm is used to inverted. (c)-For a lens between object and screen that will allow an image to be produced? llength 1o cm, what is the minimum distance ?, 15 em do: 120 cm di ? d25cm 120(15em)(20 cm) 15 di 30o0 120 m- 25cm dle 25m 12.0m(25(120cm) li dle o 120 m verted aprighi 10 cm Iu

Explanation / Answer

a)

1/v + 1/u = 1/f

here v = image distance , u = object distance , f = focal length

Now, here if the object is 120 cm from the screen and the object distance is u , so image distance is 120 - u

So, 1/(120 - u) + 1/u = 1/25 cm

Solving we get : u = 35.5 cm and u = 84.5 cm <---- these are the two locations

b)

For, u = 35.5 cm,

magnification , m = -v/u = -(120 - u) / u

= -(120 - 35.5)/35.5

= - 2.38 <----- answer

For this case , image is real and inverted

For, u = 84.5 cm,

m = (120 - 84.5)/84.5 = -0.42 <------ answer

For this case , image is real and inverted

c)

Again coming to our main equation,

1/v + 1/u = 1/f

So, distance between object and the screen = object and the image = u + v

Taking s = u + v

So, v = s - u

So, 1/(s - u) + 1/u = 1/f

So, [ u + (s-u) ] / (u*(s - u)) = 1/f

So, s/(u*(s - u)) = 1/f

So, fs = u*(s - u)

So, s(f - u) = -u^2

So, s = u^2/(u - f)

Now, differentiating s with respect to u :

ds/du = [ 1/(u -f) ]*(2u) + u^2(-1)/(u-f)^2

= 2u/(u-f) - u^2/(u-f)^2

= [ u/(u-f) ]*(2 - u/(u-f))

So, for minimum value of object and screen, s must be minimum. And thus ds/du = 0

So, taking u /= f, (2 - u/(u -f)) = 0

So, 2 = u/(u-f)

So, 2u - 2f = u

So, u = 2f <----- object must be kept at 2 times the focal length (ie at the center of curvature)

So, plugging in the values :

1/v + 1/2f = 1/f

So, 1/v = 1/2f

So, v = 2f

So, minimum distance between object and screen = u + v

= 2f + 2f = 4f

= 4*25 cm = 100 cm = 1m <------answer

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