Reflection At Spherical Mirrors

Convex mirrors are diverging, so in the Cartesian sign convention their focal length is positive, here $f=+15$ cm, while the real object gives $u=-30$ cm. Applying $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ yields $\frac{1}{v}=\frac{1}{15}-\frac{1}{-30}=\frac{1}{15}+\frac{1}{30}=\frac{2+1}{30}=\frac{1}{10}$, so $v=+10$ cm. A positive image distance means the image lies 10 cm behind the mirror, virtual and erect, which is why such mirrors give a wide, diminished field of view. A convex mirror always behaves this way because its reflecting surface curves away from the incident light, so parallel rays diverge on reflection and only their backward extensions meet behind the surface at a virtual focus. As a result, for any real object the image stays between the pole and the focus, remaining upright and shrunken—properties that let a driver survey a broad span of traffic in a compact mirror, at the cost of making vehicles appear farther away than they are. The option 30 cm behind is wrong because it equals the object distance, which a diverging mirror never reproduces. The option 6 cm in front is wrong since a convex mirror cannot form a real image of a real object. The option 15 cm behind is wrong because that is the focal point, approached only for objects at \infty. This matches the NCERT result that convex mirrors always form virtual, upright, reduced images. A magnitude check shows $|v|<|u|$, confirming the expected diminished image.