Step 1: Use the mirror formula
The mirror formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \): \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where:
- \( f \) is the focal length,
- \( v \) is the image distance,
- \( u \) is the object distance.
Step 2: Substitute the given values
Given:
- Focal length \( f = -15 \, \text{cm} \) (for concave mirror, focal length is negative),
- Object distance \( u = -10 \, \text{cm} \) (object is always placed on the same side as the incoming light).
Substitute these values into the formula: \[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-10} \] \[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{10} \] \[ \frac{1}{v} = \frac{-2 + 3}{30} \] \[ \frac{1}{v} = \frac{1}{30} \] \[ v = 30 \, \text{cm} \]
Answer:
Therefore, the image distance is \( 30 \, \text{cm} \). So, the correct answer is option (1).
\(XPQY\) is a vertical smooth long loop having a total resistance \(R\), where \(PX\) is parallel to \(QY\) and the separation between them is \(l\). A constant magnetic field \(B\) perpendicular to the plane of the loop exists in the entire space. A rod \(CD\) of length \(L\,(L>l)\) and mass \(m\) is made to slide down from rest under gravity as shown. The terminal speed acquired by the rod is _______ m/s. 
A biconvex lens is formed by using two plano-convex lenses as shown in the figure. The refractive index and radius of curvature of surfaces are also mentioned. When an object is placed on the left side of the lens at a distance of \(30\,\text{cm}\), the magnification of the image will be: 