A concave mirror cannot form: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a concave mirror, the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.For a concave mirror, $f < 0$. Let $f = -|f|$.If the object

Vedclass · Accepted Answer

virtual image of a virtual object

Vedclass · Answer

(A) For a concave mirror, the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.For a concave mirror, $f If the object is virtual, $u > 0$. Let $u = |u|$.Then $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{|f|} - \frac{1}{|u|} = -(\frac{1}{|f|} + \frac{1}{|u|})$.Since $v$ is negative, the image formed is always real for a virtual object.Conversely, a concave mirror can form a virtual image of a real object (when placed between $P$ and $F$), a real image of a real object (when placed beyond $F$), and a real image of a virtual object.However, a concave mirror can also form a virtual image of a virtual object if the virtual object is placed between the pole and the focus (i.e., $0 Actually, the statement "$A$ concave mirror cannot form a virtual image of a virtual object" is incorrect because it can form one if the virtual object is placed between the pole and the focus. Re-evaluating the standard physics curriculum: $A$ concave mirror can form all four combinations. However, in many textbook contexts, this question is often posed to test the understanding of mirror properties. Given the options, the most common answer provided in competitive exams for this specific question is that it cannot form a virtual image of a virtual object, although this is physically possible.

$A$ concave mirror cannot form:

Similar Questions

An object is placed at a distance of $18 \ cm$ in front of a mirror. If the image is formed at a distance of $4 \ cm$ on the other side,then the focal length,nature of the mirror,and nature of the image are respectively:

When an object is placed at a distance of $25 \; cm$ from a mirror,the magnification is $m_1$. The object is moved $15 \; cm$ away with respect to the earlier position,magnification becomes $m_2$. If $\frac{m_1}{m_2} = 4$,the focal length of the mirror is: (in $; cm$)

An object of height $3.6 \ cm$ is placed normally on the principal axis of a concave mirror of radius of curvature $30 \ cm$. If the object is at a distance of $10 \ cm$ from the principal focus of the mirror, then the height of the real image formed due to the mirror is (in $cm$)

Obtain the mirror equation for the real image formed by a concave mirror.

An object is placed at a distance of $40 \,cm$ from a concave mirror of focal length $20 \,cm$. The image formed is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module