2. Consider the concave glass block (n 1.5) in air (n-1.0) is shown below. The r

Question

2. Consider the concave glass block (n 1.5) in air (n-1.0) is shown below. The radius ground on the end is 10 cm. A flaw in the glass is located 30 cm inside the glass along its axis at S. Where will the image of the flaw appear to an observer on the left? What is its transverse magnification Mr? Sketch the rays and show that they qualitatively make sense with your answers. 10 cm 3. Do problem 2 again but now with the block convex. 30 cm 10 cm z 4. Do problem 2 a third time but now with the surface of the block being flat. This is like the pin in the water glass problem we did in class k- 30 cm

Explanation / Answer

gaussean formula for refractionat single surface is given by

n1/u + (n2 - n1)/R = n2/v [ where u and v are object and image distances]

2. so, u = -30 cm

n1 = 1.5

n2 = 1

R = 10 cm

-1.5/30 - 0.5/10 = 1/v

v = -10 cm

-ve siogn meaning that the image is formed to the left of the refracting surface ( that can be seen form the ray diagram)

also, magnification = v/u = 0.333

3. for the same problem

u = -30 cm

n1 = 1.5

n2 = 1

R = -10 cm

-1.5/30 + 0.5/10 = 1/v

v = infinity

as seen from the image the image is formed at infinity

magnification = v/u = infinity

4. for the same problem

u = -30 cm

n1 = 1.5

n2 = 1

R = infinity

-1.5/30 = 1/v

v = -20 cm

as seen from the image the image formed is to the left of the refracting surface

magnifivcation m = v/u = 20/30 = 0.666

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