{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

microbeproblem

# microbeproblem - n w R 2 1 s =(1-n w-R 1 s = n w-1 R-2 n w...

This preview shows page 1. Sign up to view the full content.

R/2 R n=1 n g =5/3 n w =4/3 2. A hemispherical water droplet (radius R ) rests on a glass slide (thick- ness R ). A microbe sits in the middle of the droplet a distance R/ 2 above the slide. The indices of refraction for the water and glass are n w = 4 / 3 and n g = 5 / 3, respectively. (a) Find the apparent position of the microbe when viewed from above the droplet. (b) Find the apparent position of the microbe when viewed from di- rectly below the glass slide. (First ﬁnd the image position for refraction at the water-glass interface. Then use this image as the object for refraction at the glass-air interface.) Solution (a) We can use the equation for images formed by refraction, but we note that the center of curvature of the spherical interface is on the same side as the object (microbe). By convention, this means the radius of curvature in the equation is negative:
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n w ( R/ 2) + 1 s = (1-n w )-R 1 s = ( n w-1) R-2 n w R =-(1 + n w ) R ⇒ s =-3 7 R. The image from above is virtual and is formed in the water, (3 / 7) R below the surface. (b) Using the same equation, we can ﬁrst ﬁnd the image formed by refraction at the water-glass interface. Taking R = ∞ for the ﬂat interface yields, n w ( R/ 2) + n g s = 0 ⇒ s =-n g 2 n w R =-5 8 R. This image is within the water, (5 / 8) R from the water-glass interface. It serves as the object for the image formed by refraction at the glass-air inter-face. The distance to that interface is (5 / 8) R + R = (13 / 8) R : n g (13 / 8) R + 1 s = 0 ⇒ s =-(13 / 8) n g R =-39 40 R. The image seen from below the glass is within the glass, (39 / 40) R from the glass-air interface....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online