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007 (part 1 of 4) An object is placed 24.5 cm in front of a convex mirror of foc

ID: 1386218 • Letter: 0

Question

007 (part 1 of 4)

An object is placed 24.5 cm in front of a convex mirror of focal length 50 cm.

Note:

Do convex mirrors have a positive or

negative focal length?

What is its image position?

Answer in units of cm.

008 (part 2 of 4) 10.0 points

The magnification of the image in the previ-

ous question is:

009 (part 3 of 4) 10.0 points

With the arrangement discussed above, the

image is:

1.

virtual, erect, smaller than object

2.

virtual, erect, bigger than object

3.

real, same size as object

4.

virtual, inverted, bigger than object

5.

real, erect, bigger than object

6.

real, erect, smaller than object

7.

virtual, same size as object

8.

real, inverted, bigger than object

9.

virtual, inverted, smaller than object

010 (part 4 of 4) 10.0 points

An object is placed at 66.3 cm in front of this

mirror.

What is its image position?

Answer in units of cm

Explanation / Answer


for convex   mirror, its focal length is -ve

so apply for converging ( concave) lens 1/f = 1/u + 1/v

u = oject distance,

v= image distance

and f = focal length

magnifiication m = -(v/u)


Power P of a lens = 1/focal length in meter

Power P is measured in DIopters

apply for diverginh lens (convex lens) -1/f = 1/u + 1/v

u = oject distance, v= image distance and f = focal length

magnifiication m = -(v/u) =y'/y ) =y'/y


so


part A:

1/v = 1/f -1/u

v = uf/(u-f)

v = -24.5 * 50/(24.5 + 50)

v = 16.44 cm

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

magnification m =   -(-16.44)/24.5

m = 0.671

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

6.

real, erect, smaller than object

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


if   u = 66.3 cm


v = uf/(u-f)

v = -66.3 * 50/(66.3 + 50)

v =   -28.5 cm