# Light & Mirrors

1. Apr 27, 2008

### Andy111

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

1) A 24 cd point source lamp and a 73 cd point source lamp cast equal illuminances on a wall. If the 24 cd lamp is 6.3m from the wall, how far from the wall is the 73 cd lamp? Answer in units of m.

2) Two lamps illuminate a screen equally. The first lamp has an intensity of 100 cd and is 6.7m from the screen. The second lamp is 7.7 m from the screen. What is the intensity of the second lamp? Answer in units of cd.

3) Determine the minimum height of a vertical flat mirror in which a person 67 in. in height can see his full image. Answer in units of in. (Hint: draw a ray diagram)

4)A concave mirror has a focal length of 24.8cm. Determin the object position for which the resulting image is upright and 4 times the size of the object. Answer in units of cm.

2. Relevant equations

Focal length= $$\frac{Radius}{2}$$

$$\frac{1}{focal length}$$ = $$\frac{1}{distance of object}$$ + $$\frac{1}{distance of image}$$

magnification = $$\frac{-distance of image}{distance of object}$$ = $$\frac{height of image}{height of object}$$
3. The attempt at a solution

My teacher told us we wouldn't need to know the equations for light intensity, so I have no idea how to solve 1&2.

3) I drew the ray diagram like it said and found the area of mirror needed to see the image. But obviously it's not to scale, and I doubt I could measure it and scale it well enough to get an accurate answer, because the system we use to turn in homework likes exact answers that are at most +/-1% off.

4)I tried using this $$\frac{1}{focal length}$$=$$\frac{1}{distance of object}$$+$$\frac{1}{distance of image}$$

I knew that the distance of the image was 4 times the size of the object so i substituted distance of image with 4(distance of object). Crunching the numbers I got 31, which doesn't make sense because if the object is outside the focal point of a concave mirror, then the image will appear smaller, not larger. I think I messed up because it says the size is 4 times larger, which probobly means height, not distance. But, I'm not sure what to do.

2. Apr 27, 2008

### Andy111

I found the equations for 1&2, so I don't need help with those anymore.

3. Apr 27, 2008

### alphysicist

4. The magnification is related to heights and distances by:

$$M = \frac{h_i}{h_o}= \frac{-q}{p}$$

where p is the object distance and q is the image distance. So here when you substitute you will have a minus sign (since the image is upright).

4. Apr 27, 2008

### Andy111

Thankyou so much. I got 18.6 substituting with -4 instead of 4, which was correct. I also got the 3rd question right. So I don't need any more help now.