The magnification \( m \) for a concave mirror is given by: \[ m = \frac{-v}{u} \] where \( v \) is the image distance and \( u \) is the object distance. We are given that \( m = -\frac{1}{9} \) for a real image, so: \[ \frac{-v}{u} = -\frac{1}{9} \] This implies: \[ v = \frac{u}{9} \] The mirror equation is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] For a concave mirror, the focal length \( f \) is related to the radius of curvature \( R \) by: \[ f = \frac{R}{2} = \frac{36}{2} = 18 \, \text{cm} \] Substitute \( f = 18 \) cm into the mirror equation: \[ \frac{1}{18} = \frac{1}{v} + \frac{1}{u} \] Using the relation \( v = \frac{u}{9} \), substitute this into the equation: \[ \frac{1}{18} = \frac{9}{u} + \frac{1}{u} \] Simplifying: \[ \frac{1}{18} = \frac{10}{u} \] Solving for \( u \): \[ u = 180 \, \text{cm} \] Thus, the object should be placed 180 cm from the mirror.
A hemispherical vessel is completely filled with a liquid of refractive index \( \mu \). A small coin is kept at the lowest point \( O \) of the vessel as shown in the figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point \( E \) (at the level of the vessel) is: