The double concave lens in the figure above isfabricated of glass having index o
ID: 1677805 • Letter: T
Question
The double concave lens in the figure above isfabricated of glass having index of refraction n = 1.8.The magnitudes of the radii of curvature of theleft and right surfaces of the lens are |Rl| =19 cm and |Rr| = 14 cm, respectively.
(a) Calculate the power P of the lens in"diopters".
P = diopters
An upright object of height H = 0.8 cm is placed at adistance of 19 cm to the left of the lens.
(b) Calculate the image distance.
s1' = cm
(c) Calculate the image height.
H' = cm
(d) Is the image real or virtual?
Answer as follows: Real Image = 1; Virtual Image = 2.
Image type = *
2 OK
(e) 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 5 cm to the right ofthe location of the lens, i.e. a virtualobject.
(f) Calculate the image distance in this case. (Give numericalvalue and proper algebraic sign.)
s2' = cm
(g) Is the image real or virtual?
Answer as follows: Real Image = 1; Virtual Image = 2.
Image type = *
1 OK
(h) Is the image upright or inverted?
Answer as follows: Upright = 11, Inverted = 22.
Image orientation = *
11 OK
(i) Is the image located to the right or left of thelens?
Answer as follows: Right = 111, Left = 222.
Image location = *
111 OK
Explanation / Answer
1 / f = (n - 1) (1 / r1 - 1 /r2) lens maker's formula: r is positive (from left ) if convex andnegative if concave1 / f = .8 (-1/19 - 1/14) f = -10.1 cm = -.101 m P = 1 / f = -9.9 Diopters 1 / i = 1 / f - 1 / o i = o f / (o - f) = 19 * (-10.1) / (19 + 10.1) = -6.59cm M = -i / o = 6.59 / 19 = .347 h = .8 * .347 = .278 cm virtual anderect f) o = -5 i = o f / (o - f) = -5 * (-10.1) /5.1 = 9.9 cm (to right and erect) M = -i / o = 9.9 / 5 =1.98 image is erect
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