Question:

A convex lens with a focal length of 15 cm is placed in water (refractive index of water = 1.33). A 4 cm tall object is placed 25 cm in front of the lens. Calculate the image distance, the magnification, and the nature (real or virtual) of the image formed.

Answer:

Given:

• Focal length of the lens (f) = 15 cm
• Refractive index of water (n) = 1.33
• Height of the object (h_o) = 4 cm
• Object distance (d_o) = -25 cm (for concave lens)

To find:

1. Image distance (d_i)
2. Magnification (M)
3. Nature of the image

Step 1: Calculating Image Distance (d_i)

Using the lens formula:

1/f = 1/d_o + 1/d_i

Substituting the given values:

1/15 = 1/-25 + 1/d_i

Rearranging the equation:

1/d_i = 1/15 + 1/25

Calculating the value of d_i:

1/d_i = (5 + 3)/75

1/d_i = 8/75

d_i = 75/8 cm

Therefore, the image distance (d_i) is 9.375 cm.

Step 2: Calculating Magnification (M)

The magnification (M) can be calculated using the formula:

M = -d_i / d_o

Substituting the given values:

M = -9.375 / -25

M = 9.375/25

Therefore, the magnification (M) is 0.375.

Step 3: Determining Nature of the Image

To determine the nature of the image, we use the sign convention for lenses:

• If the image distance (d_i) is positive, the image is real.
• If the image distance (d_i) is negative, the image is virtual.

From our calculation, we found that d_i = 9.375 cm, which is positive. Hence, the image formed is real.

Answer:

1. Image distance (d_i) = 9.375 cm
2. Magnification (M) = 0.375
3. Nature of the image: Real