Grade 11Mechanics

A point source of light is moving at a rate of 2 cm -s-1 towards a thin convex lens
of focal length 10 cm along its optical axis. When the source is 15 cm away from
the lens the image is moving at

[A] 4 cm-s-1 towards the lens
[B] 8 cm-s-1 towards the lens
[C] 4 cm-s-1 away from the lens [D] 8 cm-s-1 away from the lens

Shrestha Dey Sarkar

8 Years agoGrade 11

1 Answer

Askiitians Tutor Team

Approved Tutor Answer1 Year ago

To determine how fast the image is moving when the point source of light is approaching a thin convex lens, we can use the lens formula and the concept of magnification. Let's break this down step by step.

The Lens Formula

The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of the lens:

1/f = 1/v - 1/u

Here, we need to remember that the object distance (u) is taken as negative in the lens formula when the object is on the same side as the incoming light. Thus, if the object is 15 cm away from the lens, we have:

u = -15 cm

Given that the focal length (f) of the lens is +10 cm (positive for a convex lens), we can substitute these values into the lens formula.

Calculating the Image Distance

Substituting the values into the lens formula:

1/10 = 1/v - 1/(-15)

This simplifies to:

1/10 = 1/v + 1/15

To combine the fractions, we find a common denominator, which is 30:

3/30 = 1/v + 2/30

Rearranging gives:

1/v = 3/30 - 2/30 = 1/30

Thus, we find:

v = 30 cm

Understanding Image Movement

Now, we need to determine how fast the image is moving as the object approaches the lens. The object is moving towards the lens at a speed of 2 cm/s. We can use the concept of rates of change in the lens formula. Differentiating the lens formula with respect to time (t) gives us:

0 = -1/v² (dv/dt) + 1/u² (du/dt)

Here, du/dt is the speed of the object (which is -2 cm/s, since it's moving towards the lens), and dv/dt is the speed of the image we want to find.

Substituting Values

We already found that:

u = -15 cm and v = 30 cm

Now substituting these values into the differentiated equation:

0 = -1/(30)² (dv/dt) + 1/(-15)² (-2)

This simplifies to:

0 = -1/900 (dv/dt) + 1/225 (2)

Multiplying through by 900 to eliminate the fractions gives:

0 = -dv/dt + 4

Thus, we find:

dv/dt = 4 cm/s

Direction of Image Movement

Since the image distance (v) is positive and the image is moving towards the lens as the object approaches, we conclude that the image is moving towards the lens at a speed of 4 cm/s.