Question

A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image?

Answer 1

Step 1: Write down the given data.
Focal length of convex lens, \( f = +20 \, \text{cm} \) (positive for convex lens)
Object distance, \( u = -40 \, \text{cm} \) (negative as per sign convention - object on left side of lens)
Step 2: Recall the lens formula.
The lens formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( v \) is the image distance, \( u \) is the object distance, and \( f \) is the focal length.
Step 3: Substitute the values into the lens formula.
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{(-40)} \] \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \]
Step 4: Solve for \( \frac{1}{v} \).
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \] \[ \frac{1}{v} = \frac{2 - 1}{40} = \frac{1}{40} \]
Step 5: Find the image distance \( v \).
\[ v = 40 \, \text{cm} \] The positive sign indicates that the image is formed on the opposite side of the lens from the object (right side for convex lens), meaning it is a real image.
Step 6: Determine the nature of the image.

• Since \( v \) is positive, the image is real and inverted.
• Object distance \( u = -40 \, \text{cm} \) and focal length \( f = 20 \, \text{cm} \).
• Here, \( u = 2f \) (object at 2F).
• When object is at 2F, image is also at 2F on the other side, and the image size is equal to the object size.

Step 7: Calculate magnification.
\[ m = \frac{v}{u} = \frac{40}{-40} = -1 \] The negative sign confirms that the image is inverted, and \( |m| = 1 \) means the image is same size as the object.
Step 8: Final answer.
\[ \boxed{\text{Image position: } 40 \, \text{cm on the opposite side of the lens}
\boxed{\text{Nature of image: Real, inverted, and same size as the object}} \]

Answer 2

Step 1: Recall the properties of a concave lens.
A concave lens is a diverging lens. It always forms a virtual, erect, and diminished image regardless of the position of the object. The image is always formed on the same side as the object.
Step 2: Object position specified.
Here, the object is placed between the optical centre (O) and the principal focus (F) of the concave lens. This means the object is closer to the lens than the focal point.
Step 3: Ray diagram construction.

Step 4: Description of the ray diagram.

• The object (upright arrow) is placed between the optical centre (O) and the focus (F) on the left side of the concave lens.
• Ray 1: A ray parallel to the principal axis, after refraction through the concave lens, appears to diverge from the focus (F) on the same side as the object. The emergent ray is extended backward (dashed line) to meet the virtual focus.
• Ray 2: A ray passing through the optical centre goes straight without any deviation.
• The two refracted rays diverge and do not actually meet on the right side. When extended backwards (dashed lines), they appear to meet at a point on the same side as the object.
• This point of intersection gives the position of the virtual image.

Step 5: Characteristics of the image.

• Position: Between the optical centre and the focus (F) on the same side as the object, but closer to the lens than the object.
• Nature: Virtual and erect (cannot be obtained on a screen).
• Size: Diminished (smaller than the object).
• Magnification: |m|<1.

Step 6: Verification using lens formula.
For a concave lens, \( f \) is negative. Let \( f = -20 \, \text{cm} \) (example) and object distance \( u = -10 \, \text{cm} \) (between O and F). Using lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{-20} = \frac{1}{v} - \frac{1}{-10} \] \[ -\frac{1}{20} = \frac{1}{v} + \frac{1}{10} \] \[ \frac{1}{v} = -\frac{1}{20} - \frac{1}{10} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20} \] \[ v = -\frac{20}{3} \approx -6.67 \, \text{cm} \] Negative sign confirms image is on same side as object, and |v|<|u| confirms diminished image.
Step 7: Final answer.
When object is placed between O and F of a concave lens, the image formed is virtual, erect, diminished, and located between O and F on the same side as the object.

Answer 3

Step 1: Write down the given data.
Focal length of convex lens, \( f_1 = +30 \, \text{cm} \) (positive for convex lens)
Focal length of concave lens, \( f_2 = -15 \, \text{cm} \) (negative for concave lens)
Step 2: Convert focal lengths to meters for power calculation (optional, but recommended).
\[ f_1 = 30 \, \text{cm} = 0.3 \, \text{m} \] \[ f_2 = -15 \, \text{cm} = -0.15 \, \text{m} \]
Step 3: Recall the formula for equivalent focal length of lenses placed in contact.
When two lenses are placed in contact, the equivalent focal length \( F \) is given by: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 4: Substitute the values.
\[ \frac{1}{F} = \frac{1}{30} + \frac{1}{(-15)} \] \[ \frac{1}{F} = \frac{1}{30} - \frac{1}{15} \]
Step 5: Simplify the expression.
\[ \frac{1}{F} = \frac{1 - 2}{30} = \frac{-1}{30} \]
Step 6: Find the equivalent focal length.
\[ F = -30 \, \text{cm} \] The negative sign indicates that the combination behaves as a concave lens (diverging lens).
Step 7: Calculate the power of the combination.
Power of a lens (in diopters) is given by: \[ P = \frac{1}{f \, (\text{in meters})} \] First, convert \( F \) to meters: \[ F = -30 \, \text{cm} = -0.3 \, \text{m} \] \[ P = \frac{1}{-0.3} = -\frac{1}{0.3} = -\frac{10}{3} = -3.33 \, \text{D} \]
Step 8: Alternative method - Add individual powers.
Power of convex lens: \[ P_1 = \frac{1}{f_1 \, (\text{m})} = \frac{1}{0.3} = \frac{10}{3} \approx +3.33 \, \text{D} \] Power of concave lens: \[ P_2 = \frac{1}{f_2 \, (\text{m})} = \frac{1}{-0.15} = -\frac{1}{0.15} = -\frac{20}{3} \approx -6.67 \, \text{D} \] Equivalent power: \[ P = P_1 + P_2 = \frac{10}{3} + \left(-\frac{20}{3}\right) = -\frac{10}{3} = -3.33 \, \text{D} \] This matches our previous calculation.
Step 9: Final answer.
\[ \boxed{\text{Equivalent focal length } F = -30 \, \text{cm}
\boxed{\text{Equivalent power } P = -\frac{10}{3} \, \text{D} \approx -3.33 \, \text{D}} \]

Answer 4

Step 1: Write down the given data.
Focal length of concave lens, \( f_1 = -2 \, \text{m} \) (negative for concave lens)
Focal length of convex lens, \( f_2 = +1.5 \, \text{m} \) (positive for convex lens)
Step 2: Recall the formula for equivalent focal length of lenses placed in contact.
When two lenses are placed in contact, the equivalent focal length \( F \) is given by: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 3: Substitute the values.
\[ \frac{1}{F} = \frac{1}{(-2)} + \frac{1}{1.5} \] \[ \frac{1}{F} = -\frac{1}{2} + \frac{1}{1.5} \]
Step 4: Convert to common denominator.
\[ \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 \] \[ -\frac{1}{2} = -0.5 \] \[ \frac{1}{F} = -0.5 + 0.6667 = 0.1667 \]
Step 5: Calculate exactly using fractions.
\[ \frac{1}{F} = -\frac{1}{2} + \frac{2}{3} = -\frac{3}{6} + \frac{4}{6} = \frac{1}{6} \]
Step 6: Find the equivalent focal length.
\[ F = +6 \, \text{m} \] The positive sign indicates that the combination behaves as a convex lens (converging lens).
Step 7: Calculate using power method to verify.
Power of concave lens: \[ P_1 = \frac{1}{f_1} = \frac{1}{-2} = -0.5 \, \text{D} \] Power of convex lens: \[ P_2 = \frac{1}{f_2} = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 \, \text{D} \] Equivalent power: \[ P = P_1 + P_2 = -0.5 + 0.6667 = +0.1667 \, \text{D} \] Positive power confirms that the combination behaves as a convex lens.
Step 8: Reason for the behavior.
The combination behaves as a convex lens because the converging power of the convex lens (\( \frac{1}{1.5} = 0.6667 \, \text{D} \)) is greater than the diverging power of the concave lens (\( \frac{1}{2} = 0.5 \, \text{D} \)). Since \( P_2>|P_1| \), the net power is positive, resulting in a converging combination.
Step 9: Final answer.
The combination behaves as a convex lens because the power of the convex lens } \left(\frac{1}{1.5}\right) \text{ is greater than the magnitude of power of the concave lens } \left(\frac{1}{2}\right), \text{ giving a net positive power