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 \( \frac{1}{v} \).
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \] Take LCM: \[ \frac{1}{v} = \frac{2 - 1}{40} \] \[ \frac{1}{v} = \frac{1}{40} \]

Step 5: Find the image distance \( v \).
\[ v = 40 \, \text{cm} \] Since \( v \) is positive, the image is formed on the right side of the lens. Therefore, the image is real.

Step 6: Determine the nature and position of the image.
– The object distance is \( u = -40 \, \text{cm} \).
– The focal length is \( f = 20 \, \text{cm} \).
– Here, the object is placed at \( 2f \) (since \( 2f = 40 \, \text{cm} \)).
– For a convex lens, when the object is placed at \( 2F \), the image is formed at \( 2F \) on the opposite side.
– The image formed is real, inverted, and same size as the object.

Step 7: Calculate magnification.
\[ m = \frac{v}{u} \] \[ m = \frac{40}{-40} \] \[ m = -1 \] The negative sign indicates that the image is inverted, and \( |m| = 1 \) means the image is of the same size as the object.

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

Image Formation by a Concave Lens When the Object is Placed Between the Optical Centre and Principal Focus:

A concave lens is a diverging lens. It always forms a virtual, erect, and diminished image, irrespective of the position of the object.

When the object is placed between the optical centre (O) and the principal focus (F₁) of a concave lens:

– The image is formed on the same side of the lens as the object.
– The image is virtual (cannot be obtained on a screen).
– The image is erect.
– The image is diminished (smaller than the object).
– The image lies between the optical centre and the principal focus.

Ray Diagram (Conceptual Representation):

        F₁        O        F₂
---------|--------|--------|---------------- Principal Axis
          \        |        /
           \       |       /
            \      |      /
             \     |     /
              \    |    /
               \   |   /
                \  |  /
                 \ | /
                  \|/
                  )(
                  )(
                 ↑
               Object

  (Virtual image formed between O and F₁ on same side)

  

Explanation of Ray Diagram:

1. Draw one ray from the top of the object parallel to the principal axis. After refraction through the concave lens, it diverges as if it is coming from the principal focus (F₁).
2. Draw a second ray passing through the optical centre (O). This ray emerges without deviation.
3. The refracted rays diverge, but their backward extensions meet between O and F₁.
4. The point where these backward extensions meet gives the position of the virtual image.

Conclusion:
When an object is placed between the optical centre and principal focus of a concave lens, the image formed is virtual, erect, and diminished, and it appears between the optical centre and the principal focus on the same side of the lens.

Answer 3

Given:
Focal length of convex lens, \( f_1 = +30 \, \text{cm} \)
Focal length of concave lens, \( f_2 = -15 \, \text{cm} \) (negative sign for concave lens)

Step 1: Use formula for equivalent focal length of lenses in contact
For two lenses placed together: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] where \( F \) is the equivalent focal length.

Step 2: Substitute the given values \[ \frac{1}{F} = \frac{1}{30} + \frac{1}{(-15)} \] \[ \frac{1}{F} = \frac{1}{30} - \frac{1}{15} \] Take LCM: \[ \frac{1}{F} = \frac{1 - 2}{30} \] \[ \frac{1}{F} = \frac{-1}{30} \]

Step 3: Find Equivalent Focal Length \[ F = -30 \, \text{cm} \] The negative sign indicates that the combination behaves like a concave lens.

Step 4: Convert focal length into metres \[ F = -30 \, \text{cm} = -0.30 \, \text{m} \]

Step 5: Calculate Power of the Combination Power \( P = \frac{1}{F(\text{in metres})} \) \[ P = \frac{1}{-0.30} \] \[ P = -3.33 \, \text{D} \]

Final Answer: \[ \boxed{\text{Equivalent focal length} = -30 \, \text{cm}} \] \[ \boxed{\text{Power of combination} = -3.33 \, \text{D}} \]
The negative power confirms that the lens combination behaves as a diverging (concave) lens.

Answer 4

Given:
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 1: Use the formula for combination of lenses in contact
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] where \( F \) is the equivalent focal length of the combination.

Step 2: Substitute the given values \[ \frac{1}{F} = \frac{1}{-2} + \frac{1}{1.5} \] \[ \frac{1}{F} = -0.5 + 0.67 \] \[ \frac{1}{F} = 0.17 \]

Step 3: Find the equivalent focal length \[ F = \frac{1}{0.17} \] \[ F \approx 6 \, \text{m} \]
Since \( F \) is positive, the equivalent focal length is positive.

Conclusion:
The combination behaves as a convex lens.

Reason:
The positive focal length indicates that the converging power of the convex lens (1.5 m) is greater than the diverging power of the concave lens (2 m). Therefore, the overall effect is converging, and the combination behaves like a convex lens.