JAMB 2016 Physics · Question 42

If the magnification of a virtual image formed by an object 10cm from a convex less is 3, then the focal length of the lens is

Options

  1. A.10cm
  2. B.20cm
  3. C.25cm
  4. D.15cm

Answer: D. 15cm

Explanation written with AI and checked against the answer key

1. Analyze the question

We are given:

  • Object distance (uu) = 10 cm10\text{ cm}
  • Magnification (mm) = 33
  • The image is virtual.
  • We need to find the focal length (ff) of the convex lens.
2. Explain the concept

For a lens, magnification (mm) is defined as the ratio of the image distance (vv) to the object distance (uu):

m=vum = \frac{v}{u}

Since the image is virtual, the image distance (vv) must be treated as negative in the lens formula. The lens formula is:

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
3. Show the solution

Step 1: Find the image distance (vv). Using the magnification formula:

m=vum = \frac{v}{u}
3=v103 = \frac{v}{10}
v=3×10=30 cmv = 3 \times 10 = 30\text{ cm}

Because the image is virtual, we assign a negative sign to vv:

v=−30 cmv = -30\text{ cm}

Step 2: Calculate the focal length (ff) using the lens formula.

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}
1f=110+1−30\frac{1}{f} = \frac{1}{10} + \frac{1}{-30}
1f=110−130\frac{1}{f} = \frac{1}{10} - \frac{1}{30}

Find the common denominator (which is 30):

1f=3−130\frac{1}{f} = \frac{3 - 1}{30}
1f=230\frac{1}{f} = \frac{2}{30}
1f=115\frac{1}{f} = \frac{1}{15}

Therefore:

f=15 cmf = 15\text{ cm}
4. Identify the answer

The calculated focal length is 15 cm15\text{ cm}, which matches option D.

5. Tip

Always remember the Sign Convention: For thin lenses, real images have a positive image distance (vv), while virtual images have a negative image distance (vv). If you forget this sign, you will likely get the wrong focal length!

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