1) What is f, the focal length of the mirror? If the mirror is concave f is posi

Question

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.

cm

2)

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

cm

3)

The object arrow is now moved such that image distance doubles, i.e., ximage,new = -169 cm. What is yimage,new, the new y co-ordinate of the image of the tip of the arrow?

cm

4)

The object arrow is now moved to x = x1,new = -30.4 cm. What is y2,new, the new y co-ordinate of the image of the arrow?

cm

5)

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

(a)

(b)

(c)

Explanation / Answer

(x1,y1) = (-101 cm, 7.04 cm).
(x2.y2) = (-84.5 cm, ?)

(a)
Distance of object, do = 101 cm
Distance of image, di = 84.5 cm

Using thin lens eq,
1/f = 1/di + 1/do
1/f = 1/84.5 + 1/101
f = 46.0 cm
Focal length of the mirror, f = 46.0 cm
Mirror is concave !!

(b)
We know,
hi/ho = -di/do
y2/7.04 = - 84.5/101
y2 = -5.9 cm

(c)
di = 169 cm
f = 46.0 cm
1/f = 1/di + 1/do
1/46.0 = 1/169 + 1/do
do = 63.2 cm

hi/ho = -di/do
y2/7.04 = - 169/63.2
y2 = - 18.83 cm
new y co-ordinate of the image of the tip of the arrow , y2 = - 18.83 cm

(d)
do = 30.4 cm
f = 46.0 cm
1/f = 1/di + 1/do
1/46.0 = 1/di + 1/30.4
di = - 89.6 cm

hi/ho = -di/do
y2/7.04 = 89.6/30.4
y2 = 20.75 cm

the new y co-ordinate of the image of the arrow, y2 = 20.75 cm

(5)

Ray (b) is not reflected.

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