Physics Help Forum Lenses

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 Jun 17th 2010, 12:35 AM #1 Junior Member   Join Date: Jan 2010 Posts: 14 Lenses Given a convex lens of focal length of (x+5) cm and a concave lens of focal length x cm. The 2 lenses are placed 30 cm apart coaxially i.e along the same axis with the convex lens on the left while the concave lens is on the right. A light bulb is placed to the left of the convex lens at a distance of 10 cm. It is observed that light rays that emerges from the concave lens are parallel to each other. Find the value of x. Since the light rays that emerges from the concave lens are parallel,the light rays that emerges from the convex lens converges at the focal point of the concave lens. Therefore, 1/(x+5) = 1/10 + 1/(30-x) x^2-45x+100=0 Solving for x, x=2.344 or 4.266 But,the answer given is x=2.655. Part of the solution given involves the equation 1/(x+5) = 1/10 + 1/(30+x) instead of 1/(x+5) = 1/10 + 1/(30-x) They've added a negative sign to the focal length of the concave lens i.e. -x instead of x. Is the solution correct? I feel the solution is wrong because we shouldn't add a negative sign to x (the focal length of the concave lens) since x is a variable. If it's a variable shouldn't the negative sign take care of itself? E.g. x^2+6x+5=0 solving, (x+1)(x+5)=0 x=-1,-5 But,if I know x is negative (as in the case of the focal length of the concave lens), I couldn't simply add a negative sign in front of x : (-x)^2+6(-x)+5=0 x^2-6x+5=0 (x-1)(x-5)=0 x=1,5 (a different set of solution) So is the solution correct?
 Jun 17th 2010, 05:33 AM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 Converging rays from the convex lens diverge from the concave lens before actually converging to a point and emerge as parallel rays. It happens when the converging point happens to be the focal point of the concave lens. So the image distance of the convex lens becomes (30 + x).
Jun 17th 2010, 07:00 AM   #3
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 Originally Posted by sa-ri-ga-ma Converging rays from the convex lens diverge from the concave lens before actually converging to a point and emerge as parallel rays. It happens when the converging point happens to be the focal point of the concave lens. So the image distance of the convex lens becomes (30 + x).
I'm sorry but I don't seem to get your point. Can you please explain further?

Where exactly is the focal point of the concave lens at which the rays from the convex lens converge? Is the focal point you are referring to situated to the right of the concave lens or in between the convex and concave lens?
If you say the image distance of the convex lens is (30+x) cm, are you saying the focal point of the concave lens at which the rays from the convex lens converge located to the right of the concave lens?

Jun 17th 2010, 09:20 AM   #4
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 Originally Posted by cyt91 I'm sorry but I don't seem to get your point. Can you please explain further? Where exactly is the focal point of the concave lens at which the rays from the convex lens converge? Is the focal point you are referring to situated to the right of the concave lens or in between the convex and concave lens? If you say the image distance of the convex lens is (30+x) cm, are you saying the focal point of the concave lens at which the rays from the convex lens converge located to the right of the concave lens?
Yes. It is situated to the right of the concave lens. If a parallel beam of light falls on a concave lens, after refraction from the lens, appears to be diverging from a point, which is the focal point of the concave lens.
In this problem, the direction of the rays are reversed.

Jun 19th 2010, 03:32 AM   #5
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 Originally Posted by sa-ri-ga-ma Yes. It is situated to the right of the concave lens. If a parallel beam of light falls on a concave lens, after refraction from the lens, appears to be diverging from a point, which is the focal point of the concave lens. In this problem, the direction of the rays are reversed.
(?) Is it possible for the rays to reverse direction? I thought rays should travel in a straight line i.e. directed away from the light source. If the rays reverse direction,do you mean that we see no light emerging from the right side of the concave lens?

Can it be that the emergent rays from the convex lens converge at the focal point of the concave lens between the convex lens and concave lens and then emerge as parallel rays after passing the through the concave lens?
In this case,could we add a negative sign to the focal length of the concave lens i.e. -x cm. (Please refer to my first post)

Thanks a lot for your explanation.

 Jun 19th 2010, 03:42 AM #6 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 Can it be that the emergent rays from the convex lens converge at the focal point of the concave lens between the convex lens and concave lens and then emerge as parallel rays after passing the through the concave lens? No. It happens when the two lenses are convex. Actually the emergent rays from the convex lens enter the concave lens before converge at point and after refraction through the concave lens, the converging rays may converge farther point or diverge depending on the position of the concave lens. If the converging point happens to be the focal point of the concave lens, then the diverging rays from the concave lens are parallel. Last edited by sa-ri-ga-ma; Jun 19th 2010 at 03:47 AM.
Jun 19th 2010, 08:18 AM   #7
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 Originally Posted by sa-ri-ga-ma Can it be that the emergent rays from the convex lens converge at the focal point of the concave lens between the convex lens and concave lens and then emerge as parallel rays after passing the through the concave lens? No. It happens when the two lenses are convex. Actually the emergent rays from the convex lens enter the concave lens before converge at point and after refraction through the concave lens, the converging rays may converge farther point or diverge depending on the position of the concave lens. If the converging point happens to be the focal point of the concave lens, then the diverging rays from the concave lens are parallel.
I understand your point. Just looked up a reference book. Thanks a lot! You've been very helpful. God bless you.

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