Question 5, chap 136, sect 6. part 1 of 2 10 points A lens of thickness t = 8.4

Question

Question 5, chap 136, sect 6. part 1 of 2 10 points A lens of thickness t = 8.4 cm has convex radii of curvature of magnitude R1 = 9.8 cm and R216.8 cm. The index of refrac- tion of the lens is n = 1.56. An object "O" is placed 37.8 cm to the left of the lens. Ri R2 Where is 11, the image due to the first surface, measured from that surface (with positive to the right)? Answer in units of cm. Question 6, chap 136, sect 6. part 2 of 2 10 points Where is 12, the image due to the second surface, relative to the first surface? Answer in units of cm.

Explanation / Answer

for first surface

object distance s1 = 37.8 cm

radius of curvature R1 = 9.8 cm

1/s1 + n/s1' = (n-1)/R1

1/37.8 + 1.56/s' = (1.5-1)/9.8

s1' = +63.504 cm   <<<<==============ANSWER

===================================

for right surface

object distance s2 = -(63.504-8.4) = -55.104 cm

R2 = -16.8 cm

n/s2 + 1/s2' = (1-n)/R2

-1.56/55.104 + 1/s2' = -(1-1.56)/16.8

s2' = 16.22 cm <<<<==============ANSWER

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.