# Homework Help: Geometric Optics

1. Jun 16, 2010

### cyt91

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

$$\frac{1}{f}$$=$$\frac{1}{u}$$+$$\frac{1}{v}$$

3. The attempt at a solution
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,

$$\frac{1}{x+5}$$=$$\frac{1}{10}$$+$$\frac{1}{30-x}$$
$$x^{2}$$-45x+100=0

Solving for x, x=2.344 or 4.266

Part of the solution given involves the equation

$$\frac{1}{x+5}$$=$$\frac{1}{10}$$+$$\frac{1}{30+x}$$

instead of $$\frac{1}{x+5}$$=$$\frac{1}{10}$$+$$\frac{1}{30-x}$$

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.

2. Jun 16, 2010

### Mindscrape

Yes, that is right, not only if you were to go through the geometric math of it, but you should also be able to simply think about it. A convex lens takes parallel light and converges it, while a concave lens takes parallel light and diverges it. They behave opposite of one another.

Basically it means you used the wrong focus point by making the focus on the the wrong side of the concave lens.

3. Jun 17, 2010

### cyt91

But 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?

4. Jun 17, 2010

### aim1732

Yes the solution is correct.
The variable x given is actually a modulus value - always positive. Think of it this way - if x were the variable you were saying it could be positive or negative but not both at the same time. Then the lenses would either be concave or convex but not both.
Actually x is distance and by definition a +ve quantity so when we say a concave lens has focal length x it is actually -x.

5. Jun 17, 2010

### cyt91

Ok. I get your point. Thanks a lot. This is very helpful.