91Ó°ÊÓ

Object distance p = 10cm

Radius of curvature r =6cm

Refractive index of glass n₂=1.5

We use the mirror equation related to the image distance, object distance and the refractive index. Using this relation, we can find the distance between the object and the image. As the object distance increase further, the image distance increases and this will give us the range of interval for the virtual image.

Formulae:

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$

The object distance p, the image distance i and the radius of curvature r of the surface are related as

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$

Where, n₁ is the index of refraction of the material where the object is located and n₂ is the index of refraction on the other side of the surface.

As the surface faced by the object is convex, r is positive.

n₁ = 1 is the refractive index of air.

n₂ =1.5 is refractive index of the glass.

$$\begin{gathered} \frac{1}{10cm}+\frac{1.5}{i}=\frac{1.5-1}{6.0cm} \\ \frac{1.5}{i}=\frac{0.50}{6.0cm}-\frac{1}{10cm} \\ i=-(60×1.5cm) \\ =-90cm \end{gathered}$$

Thus, the distance between the object and the image is

$$\begin{gathered} d=-90\mathrm{cm}+10\mathrm{cm} \\ =-80cm \end{gathered}$$

Hence, the distance between the object and the image formed by the glass rod is 80cm.

From equation (1), we can write

As we increase the image distance p from small values to p₀,

$\frac{1}{p_0}+\frac{1.5}{i}=\frac{15-1}{r}$

$i→-∞$hence, the second term on the left-hand side will be zero as we know

$\frac{1}{-∞}=0$

$$\begin{gathered} \frac{1}{p_0}=\frac{15-1}{r} \\ {p}_0=2r \end{gathered}$$

p₀=2*6.0cm

=12cm

Thus, the given interval of range for producing the virtual image is $0<p≤12cm$.