Question

One end of a long glass rod (n=1.5)is a convex surface of radius 6.0cm.An object is located in air along the axis of the rod, at a distance of 10cmfrom the convex end (a) How far apart are the object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

Short answer

1. Distance between the object and the image formed by the glass rod is 80cm.
2. The given interval of the range for producing the virtual image is $0<p\le 12cm.$

Step-by-step solution

Listing the given quantities

Object distance p = 10cm

Radius of curvature r =6cm

Refractive index of glass n₂=1.5

Understanding the concepts mirror equation

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:

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

(a) Calculations of the distance between the object and the image

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.

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

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.

(b) Calculations of the range of distances from the end of the rod 

From equation (1), we can write

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

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

$i\to -\infty$hence, the second term on the left-hand side will be zero as we know

$\frac{1}{-\infty }=0$

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

We know r = 6.0

p₀=2*6.0cm

=12cm

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