What was the focal distance (f) when the radius of curvature was 0.70 m and inde

Question

What was the focal distance (f) when the radius of curvature was 0.70 m and index of refraction was 1.8? _____________ 6. Calculate the radius of curvature of a lens with a focal distance of 40. cm and an index of 1.2. (WATCH YOUR UNITS) ____________________ 7. An object placed 35cm away from a lens projects a real image 0.55m behind the lens. What is this lens’ focal distance? _____________________ 8. What is the lens’ magnification? ____________________ 9. An object 20. cm to the left of a convex lens is 1.0 m in height. What is the height and location of its image if the lens has a magnification of -2.0? ________________ m and ________________ cm on the (left / right) side of the lens 10. Imagine you are nearsighted (can only see close objects clearly and far objects are blurry). At your last eye doctor’s appointment, your optometrist tells you that she will need to increase your prescription because as it turns out light is focusing “too soon” or in front of your retina. You respond, “Obviously, I will need a (converging / diverging) lens with a (longer / shorter) focal distance.”

Explanation / Answer

5) f= focal length = ?

R = radius of curvature = 0.70 m

u= index of refraction = 1.8

from lens maker formula :[ here R1 = R2 = R ]

1/f = (u-1)*2/R

1/f = ( 1.8 - 1)*2/0.70

f = 0.4375 m

-------------------------

6) R= radius of curvature = ?

f = focal length = 40 cm

u= index of refraction = 1.2

1/f = (u-1)*2/R

1/40 = (1.2-1)*2/R

R = 16 cm

------------------------------

7) u = object distance = - 35 cm

v= image distance = 0.55 m = 55 cm ( +ve because of real image)

f = focal length = ?

1/f = 1/v - 1/u

1/f = 1/ 55 - 1/ -35

f= 21.38 cm

-----------------------------

8) m= linear magnification = ?

m= v/- u

m= 55 /-( -35 )

m= 1.57

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

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