A spherical mirror forms an image of an arrow at x = x 2 = -87.7 cm. The mirror

Question

A spherical mirror forms an image of an arrow at x = x2 = -87.7 cm. The mirror is represented in the drawing as a plane at x = 0. The tip of the object arrow is located at (x,y) = (x1,y1) = (-106 cm, 7.08 cm).

1.What is f, the focal length of the mirror? If the mirror is concave f is positive. If the mirror is convex, f is negative.

2.What is y2, the y co-ordinate of the image of the tip of the arrow?

3.The object arrow is now moved such that image distance doubles, i.e., ximage,new = -175.4 cm. What is yimage,new, the new y co-ordinate of the image of the tip of the arrow?

4.             The object arrow is now moved to x = x1,new = -32.2 cm. What is y2,new, the new y co-ordinate of the image of the arrow?

5.Which of the rays in the diagram is not reflected as shown?

A spherical mirror forms an image of an arrow at x = x2 = -87.7 cm. The mirror is represented in the drawing as a plane at x = 0. The tip of the object arrow is located at (x,y) = (x1,y1) = (-106 cm, 7.08 cm). 1.What is f, the focal length of the mirror? If the mirror is concave f is positive. If the mirror is convex, f is negative. 2.What is y2, the y co-ordinate of the image of the tip of the arrow? 3.The object arrow is now moved such that image distance doubles, i.e., ximage,new = -175.4 cm. What is yimage,new, the new y co-ordinate of the image of the tip of the arrow? 4. The object arrow is now moved to x = x1,new = -32.2 cm. What is y2,new, the new y co-ordinate of the image of the arrow? 5.Which of the rays in the diagram is not reflected as shown?

Explanation / Answer

Given that,

image dist =i = -87.7 cm ; object dist. = o = -106cm (check the coordinates given (x1,y1) = (-106 cm, 7.08 cm).

(1)Let f be the focal length.

From lens formula we have,

1/f = 1/i + 1/o

f = o * i / o + i = (-106 )x (-87.7) / (-106 -87.7) = - 9296.2/193.7 = -47.99 = -48 cm

But because the focal point is at the same side where the object is, so the mirror is concave, so we will ignore the negative sign.

Hence , f = 48 cm

(2)for finding y2, we will use (x1,y1) = (-106 cm, 7.08 cm).

M = h'/h = -i/o

y2 = h' = - i x h / o = -(-87.7) (7.08) / (-106) = -5.85 cm

(3)new image dist = i' = -175.4 cm and we have f= -48 cm, we have to find o in order to find the new coordinates

o = i' x f / i - f = (-175.4) (-48) / -175.4 - (-48) = -8419.2 / 127.4 = -66.08 cm

Again, h'/h = -i'/o

h' = - i' x h / o = - (-175.4) (7.08) / (-66.08) = -18.79 cm

(4) o' = -32.2 cm ; f = -48 and i

i = o f /(o -f ) = (-32.2) (-48) / (-32.2 - (-48) = 1545.6/16 = 96.6 cm

Again,

h'/h = -i/o' gives h' = -i x h /o

h' = -(96.6)(7.08)/ (-32.2) = 21.24 cm

(5) ray (b) is the answer

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