Lenses~focal length, object & image distance

[vickilcw]

1)Question:
The focal length of a convex lens is 16cm.
Where must be an object be placed so that the image formed by the lens is
a)real and twice as large
b)virtual and twice as large

2)Related equation:
i)magnification=(height of image) over (height of object)
ii)magnification=(image distance) over (object distance)

3)Attempt to this question:
a)As the manification is 2,the distance between the image and the lens (let this
distance be i cm)is twice of that between the object and the lens (let this distance
be b cm).
Thus, i=2b
As 2F'>b>F',therefore, 32>b>16
and i>2F, thus, 2b>32 (i.e.b>16)
To conclude, 32>b>16,and hence, the object should be placed 16to32 cm from the
lens.
b)The object distance (let it be a cm) is less than F', thus, a<16.

I don't know how to calculate the exact number of object distance in these 2 questions.

 

[timeforplanb]

f-focal length
s-object distance
s'-image distance
M-magnification

a. To form a real image twice as big as the object, we let the magnification be M=-2 (check out the sign convention at the bottom in case you don't know yet)
But, M=-(s'/s), and since we're solving for s, we manipulate the eqn above to get s'=-sM. Now, substitute that to the lens eqn, (1/f)=(1/s)+(1/s'), and you'll get (1/f)=(1/s)-(1/sM). Simplifying, we get s=[f(M-1)]/sM

There you go :D All you have to do next is to substitute all the given values to the equation and you'll get s=24 cm. For part b, you just have to use M=2, and the answer's s=8 cm. Notice that both answers are positive - this is because we're dealing with a real object.

Hope this helps you :D

*Sign conventions*
f: (+) for converging/convex lens, (-) for diverging/concave lens.
s: (+) if object is on the same side of the lens as the incident light (real object), (-) if the object is on the side of the lens opposite the incident light (virtual object)
s': (+) if the image is on the side of the lens opposite the incident light (real image), (-) if the image is on the same side as the incident light (virtual image)
M: (+) if the image has the same orientation as the object (virtual image), (-) if the image is opposite to the orientation of the object (real image)

 

[PeterO]

f-focal length
s-object distance
s'-image distance
M-magnification

a. To form a real image twice as big as the object, we let the magnification be M=-2 (check out the sign convention at the bottom in case you don't know yet)
But, M=-(s'/s), and since we're solving for s, we manipulate the eqn above to get s'=-sM. Now, substitute that to the lens eqn, (1/f)=(1/s)+(1/s'), and you'll get (1/f)=(1/s)-(1/sM). Simplifying, we get s=[f(M-1)]/sM

There you go :D All you have to do next is to substitute all the given values to the equation and you'll get s=24 cm. For part b, you just have to use M=2, and the answer's s=8 cm. Notice that both answers are positive - this is because we're dealing with a real object.

Hope this helps you :D

*Sign conventions*
f: (+) for converging/convex lens, (-) for diverging/concave lens.
s: (+) if object is on the same side of the lens as the incident light (real object), (-) if the object is on the side of the lens opposite the incident light (virtual object)
s': (+) if the image is on the side of the lens opposite the incident light (real image), (-) if the image is on the same side as the incident light (virtual image)
M: (+) if the image has the same orientation as the object (virtual image), (-) if the image is opposite to the orientation of the object (real image)

Nice of you to solve the problem for OP! Perhaps you could have asked whether OP was familiar with the lens formula(e).