Question

An object is placed 30 cm from a thin convex lens of focal length 10 cm. The lens forms a sharp image on a screen. If a thin concave lens is placed in contact with the convex lens, the sharp image on the screen is formed when the screen is moved by 45 cm from its initial position. Calculate the focal length of the concave lens.

Answer 1

Calculation of Focal Length of Concave Lens

Given:

• Object distance from convex lens, \( u = -30 \, \text{cm} \)
• Focal length of convex lens, \( f_1 = +10 \, \text{cm} \)
• Shift in image position with concave lens added, \( \Delta v = 45 \, \text{cm} \)

Step 1: Use Lens Formula for Convex Lens

The lens formula is:

\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} \]

Substituting the given values for \( f_1 = 10 \, \text{cm} \) and \( u = -30 \, \text{cm} \):

\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15} \]

So, the image distance \( v \) is:

\[ v = \frac{15}{2} = 7.5 \, \text{cm} \]

The initial image distance is \( 15 \, \text{cm} \) (since the image formed is real, we use object-image symmetry).

Step 2: Adding the Concave Lens

After adding the concave lens, the image shifts by \( 45 \, \text{cm} \) behind the original screen. So, the new image distance becomes:

\[ v' = 15 + 45 = 60 \, \text{cm} \]

Step 3: Effective Focal Length of the System

Since the lenses are in contact, the effective focal length \( f \) of the system can be calculated using the lens formula:

\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]

Now, use the lens formula again for the system:

\[ \frac{1}{v'} - \frac{1}{u} = \frac{1}{f} \]

Substituting the values \( v' = 60 \, \text{cm} \) and \( u = -30 \, \text{cm} \):

\[ \frac{1}{60} + \frac{1}{30} = \frac{1}{f} = \frac{1}{20} \]

Hence, the effective focal length is \( f = 20 \, \text{cm} \).

Step 4: Focal Length of the Concave Lens

Now, we can solve for the focal length \( f_2 \) of the concave lens: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \] Substituting the known values: \[ \frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \] Solving for \( f_2 \): \[ \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \] Therefore, the focal length of the concave lens is: \[ f_2 = -20 \, \text{cm} \]

Final Answer:

The focal length of the concave lens is \( f_2 = -20 \, \text{cm} \).