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

Focal Length Determination for a Concave Lens

Provided Data:

• Object distance from a convex lens: \( u = -30 \, \text{cm} \)
• Focal length of the convex lens: \( f_1 = +10 \, \text{cm} \)
• Image position shift upon addition of a concave lens: \( \Delta v = 45 \, \text{cm} \)

Procedure 1: Convex Lens Image Formation

Utilize the lens equation:

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

Inputting \( f_1 = 10 \, \text{cm} \) and \( u = -30 \, \text{cm} \):

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

The image distance \( v \) is calculated as:

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

The initial image distance is \( 15 \, \text{cm} \), assuming a real image formation and considering symmetry.

Procedure 2: Introduction of the Concave Lens

The image is displaced by \( 45 \, \text{cm} \) further away from the original screen subsequent to the concave lens insertion. The new image distance is:

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

Procedure 3: Combined System Focal Length

For lenses in contact, the effective focal length \( f \) is given by:

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

Apply the lens equation to the combined system:

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

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

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

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

Procedure 4: Concave Lens Focal Length Calculation

Determine the focal length \( f_2 \) of the concave lens using the effective focal length formula:

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

Insert known values:

\[ \frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \]

Solve for \( f_2 \):

\[ \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \]

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

Conclusion:

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