An object is placed at a distance p = 25 cm from a concave mirror of focal length f = 10 cm. About the mirror is poured a small amount of liquid whose refractive index n = 1.4. Determine the distance of the image formed by this optical system to the mirror.
If t is the thickness/height of liquid shift due to refraction = (n-1)t=0.4t=s
First refraction takes place, then reflection then again refraction.
For reflection 1/u+1/v=1/f gives, 1/(25+s)+1/v=1/10, you can find v from this.
Final position of image is obtained by subtracting shift due to final refraction and is v-s. You can try to do away with t by approximation (t is small) as it is not given.
Hello,
Thanks to answer me, but I don't understand your solution.
1 - "(n-1)t = s" ===> What is it ? How did you know that ?
Do you know how to proof it ?
2 - What is v ?
3 - Why v - s is the answer?
Sorry, i'm confused.
1. This is a direct forumla for shift. You can also get it from real depth/apparent depth formula. Shift is the distance between object location and image location (just like the bottom of the pool appears closer by s).
2. v is the image distance due to reflection by mirror alone.
3. v is the image distance from the mirror but since there is another refraction due to water this image wil be shifted by s. Thus v-s.