The double convex lens in the figure above isfabricated of glass with index of r
ID: 1677804 • Letter: T
Question
The double convex lens in the figure above isfabricated of glass with index of refraction n = 1.7. Themagnitudes of the radii of curvature of the leftand right surfaces of the lens are |Rl| = 18 cmand |Rr| = 13 cm, respectively.
(a) Calculate the power P of the lens in"diopters".
P = diopters
An upright object of height H = 0.5 cm is placed at adistance of 22 cm to the left of the lens.
(b) Calculate the image distance. (Give a numerical value andproper algebraic sign.)
s1' = cm
(c) Calculate the magnitude of the image height.
H' = cm
(d) Is the image real or virtual?
Answer as follows: Real Image = 1; Virtual Image = 2.
Image type = *
1 OK
(e) Is the image upright or inverted?
Answer as follows: Upright = 11, Inverted = 22.
Image orientation = *
22 OK
The object is now moved to a new location, a distance 8 cm tothe left of the lens.
(f) Calculate the new image distance. (Give numerical value andproper algebraic sign.)
s2' = cm
(g) Is the image real or virtual?
Answer as follows: Real Image = 1; Virtual Image = 2.
Image type = *
2 OK
(h) Is the image upright or inverted?
Answer as follows: Upright = 11, Inverted = 22.
Image orientation = *
11 OK
The object is removed and instead converging rays are incomingfrom the left. In the absence of the lens, these rays would form anupright image at a position 18 cm to the right ofthe location of the lens, i.e. a virtualobject.
(i) Calculate the image distance in this case. (Give numericalvalue and proper algebraic sign.)
s3' = cm
(j) Is the image real or virtual?
Answer as follows: Real Image = 1; Virtual Image = 2.
Image type = *
1 OK
(k) Is the image upright or inverted?
Answer as follows: Upright = 11, Inverted = 22.
Image orientation = *
11 OK
(l) Is the image located to the right or left of thelens?
Answer as follows: Right = 111, Left = 222.
Image location = *
111 OK
Explanation / Answer
given that n = 1.7 R1 = 18cm R2 = 13cm from the lens makers formula 1/f = (n-1) ( 1/R1 - 1/R2) = 1-1.7 (1/18cm - 1/13cm ) = ------cm power = p = 1/f = -------diapoters b)from the lens formula 1/d0 + 1/di = 1/f b) h0 = 0.5cm , d0 = 22cm 1/di = 1/f - 1/d0 plug in values di = -----cm c) m =- di/do plug in values m = ------ image height = hi = mh0 plug in values hi = ------cm now object distance = d0=s11 =22cm + 8cm = 30cm new image distance 1/s21 = 1/f -1/s11 plug in values s21 =-------cm
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