I dont understand this. Also v-u = 2 m (u is negative as object distance is take

Question

I dont understand this. Also v-u = 2 m (u is negative as object distance is takenleftwards from lens ) FROM INFO BELOW:
A 2.0-cm-diameter spider is 2.0 m from a wall.

Determine the focal length and position (measured from thewall) of a lens that will make a half-size image of the spider
on the wall.

Let the distance of wall from lens be v and distance of spider fromlens be u .
Here it is given that image size id half as that of object size.
Also as we want the image on the wall , so it must be real andhence inverted ,
So v/u = -1/2
=>u=-2v
Also v-u = 2 m (u is negative as object distance is takenleftwards from lens ) <--------
Putting the values
3v =2 m
So v = 0.67 m
so u = -1.33 m

Putting the values in lens formula


=>=0.44 m
So a convex lens of focal length 0.44 m should be placed at adistance 0.67 m from the wall

Explanation / Answer

ok, so I will first define some variables:

s = distance of object from lens
s' = distance of lens from image (the wall)
h = height of object
h' = height of image
f = focal length

Instead of attempting crude, text-based diagrams, I will refer you to the ray diagrams on hyperphysics, and use two theorems:
h/s = h'/s' (1)
1/s + 1/s' = 1/f (2)
and stating the constraints given in the problem:
h' = h/2 (3)
s + s' = 2 m (4)
h = 0.01 m (5) (arbitrary variable)

now, just look at it like a long algebra problem

key to deriving (1,2) is a set of good pictures.

using (3,4,5), (1) becomes: h/s = h/[2(2-s)] ------> s = 4/3 m , and s' = 2/3 m   (6)

with (6), (2) becomes: f= (3/4 + 3/2) ^ -1 = 4/9 m

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