Questions: Spherical Mirrors: Focal Length and Image Formation

Question 1

An object is placed 5 cm from a concave mirror with focal length 10 cm. Using the mirror equation, what type of image forms and where?

• A. A real image 10 cm in front of the mirror — the focal point is the natural image location.
• B. A real image between the mirror and the focal point.
• C. A virtual image 10 cm behind the mirror — the negative image distance places it behind the surface.
• D. No image forms because the object is inside the focal length.

Answer: (C)

Using 1/s_i = 1/f − 1/s_o = 1/10 − 1/5 = −1/10, so s_i = −10 cm. The negative value means the image is virtual and located 10 cm behind the mirror. When the object is closer than the focal length (s_o < f), reflected rays diverge and never converge in front of the mirror — tracing them backward places a virtual, upright, magnified image behind the surface. This is exactly how a makeup mirror works. Option D is a common misconception: an image always forms, its nature just changes.

Question 2

Without calculating, can you determine whether a convex mirror will ever form a real image for an object placed in front of it?

• A. Yes — if the object is far enough away, the image becomes real.
• B. Yes — if the object is placed between the mirror and its focal point, a real image forms.
• C. No — a convex mirror has a negative focal length, so s_i is always negative regardless of object distance.
• D. Only if the object is placed exactly at the center of curvature.

Answer: (C)

A convex mirror has f < 0 (focal point behind the mirror). In the equation 1/s_i = 1/f − 1/s_o, with f < 0 and s_o > 0 (real object in front), 1/f is negative and 1/s_o is positive, so their difference is always negative — meaning s_i is always negative. A negative image distance always means a virtual image behind the mirror. No object position changes this. This is why convex mirrors always produce diminished, upright, wide-field virtual images used in security mirrors and car side mirrors.

Question 3

A concave mirror can form either a real or a virtual image of the same object, depending on where the object is placed relative to the focal point.

• T. True
• F. False

Answer: True

True. When the object is beyond the focal point (s_o > f), reflected rays converge in front of the mirror, forming a real, inverted image. When the object is inside the focal point (s_o < f), reflected rays diverge and the image is virtual, upright, and magnified — located behind the mirror. The focal point is the dividing line. A concave makeup mirror places your face inside the focal length; a concave solar concentrator places the target beyond it.

Question 4

A convex mirror can form a real image if the object is placed far enough away from the mirror.

• T. True
• F. False

Answer: False

False. A convex mirror has a negative focal length, which guarantees s_i < 0 for any real object (s_o > 0). No matter how far the object is, the mirror equation gives a negative image distance — always virtual, always behind the mirror. This is the misconception the Common Misconceptions section flags directly: convex mirrors form only virtual, upright, diminished images regardless of object position.

Question 5

Using the mirror equation, explain why a concave mirror forms a virtual image when the object is placed closer than the focal length.

Model answer: When s_o < f, the term 1/s_o > 1/f, so 1/s_i = 1/f − 1/s_o is negative, giving s_i < 0. A negative image distance means the image is on the same side as the incoming light — behind the mirror surface — where no real reflected rays converge. Physically, when the object is inside the focal length, the concave surface cannot bend the reflected rays enough to make them converge in front of the mirror; they remain divergent. Tracing these diverging rays backward locates a virtual image from which they appear to originate, behind the mirror.

The sign of s_i is the key diagnostic: positive means real (light actually converges there in front of the mirror), negative means virtual (reflected rays only appear to come from that point behind the mirror). The mirror equation algebraically encodes the geometric transition that occurs when the object crosses the focal point.