Question

Images Produced by Different Concave Mirrors Conceptual Question

The same object is placed at different distances d in front of six different concave spherical mirrors. Each mirror has the focal length f listed below.

Part A

Which, if any, of these scenarios produce a real image? Which, if any, of these scenarios produce a virtual image?

Sort the following options (A-F) as either real image or virtual image:

ANSWER:

A) d=15cm, f=5cm

B) d=5cm, f=20cm

C) d=10cm, f=20cm

D) d=20cm, f=10cm

E) d=10cm, f=5cm

F) d=15cm, f=20cm

Part B

Which, if any, of these scenarios produce an inverted image? Which, if any, of these scenarios produce an upright image?

Sort the following options (A-F) as either inverted image or upright image:

ANSWER:

A) d=15cm, f=5cm

B) d=5cm, f=20cm

C) d=10cm, f=5cm

D) d=15cm, f=20cm

E) d=10cm, f=20cm

F) d=20cm, f=10cm

Part C

Rank the images on the basis of the magnitude of their magnification.

Rank the following options (A-F) from largest to smallest. Some may be ranked as equal.

ANSWER:

A) d=15cm, f=5cm

B) d=15cm, f=20cm

C) d=20cm, f=10cm

D) d=5cm, f=20cm

E) d=10cm, f=5cm

F) d=10cm, f=20cm

Answer 1

Part A)

Remember if object distance is less that focus length then the image wil be vitual . And object distance is greater than focal length the image will be real.

So

(B) ,(C) & (f) have virtual image.

(A),(D) & (E) have real image.

part B)

in the concave mirror if images is virtual will be upright.

and image is real it will be inverted.

so

(A) ,(C) & (F) have inverted image.

(B), (D) &(E) have upright image.

part C)

magnification =-d_i/d

here d_i=fd/d-f (from mirror equation.)

so m=-fd /(d-f)×d =-f/d-f

here d has negative value , f is negative

so m=-f/-f+d=f/d-f

I am ignoring - sign of m​​​​​​

A) m=5/10=0. 5 , B ) m=20/5 =4, C) =10/10 =1

d) m=20/15 =1.33 , E) m=5/5 =1, F) m=20/10 =2

so

B, F, D, E=C , & A