Logo
Jeeify
MCQEasyJEE 2024Refraction & Lenses

JEE Physics 2024 Question with Solution

In an experiment to measure the focal length ff of a convex lens, the magnitude of object distance xx and image distance yy are measured with reference to the focal point of the lens. The yy-xx plot is shown in the figure. The focal length of the lens is:

  • A

    15 cm15 \, \text{cm}

  • B

    18 cm18 \, \text{cm}

  • C

    20 cm20 \, \text{cm}

  • D

    25 cm25 \, \text{cm}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The magnitudes of object distance xx and image distance yy are measured from the focal points of a convex lens.

Find: The focal length ff of the lens.

Since the distances are measured with reference to the focal points, use Newton's lens formula:

xy=f2xy = f^2

From the graph, point CC gives:

x=20 cm,y=20 cmx = 20 \, \text{cm}, \qquad y = 20 \, \text{cm}

Substituting in Newton's formula:

f2=xy=20×20=400f^2 = x y = 20 \times 20 = 400

Therefore,

f=400=20 cmf = \sqrt{400} = 20 \, \text{cm}

So, the focal length of the lens is 20 cm20 \, \text{cm}. Hence, the correct option is C.

Common mistakes

  • Using the usual lens formula directly with distances measured from the focal points. That is incorrect because here xx and yy are not measured from the optical centre. Use Newton's formula xy=f2xy = f^2 instead.

  • Reading a wrong point from the graph. The solution uses point CC with x=20 cmx = 20 \, \text{cm} and y=20 cmy = 20 \, \text{cm}. Incorrect coordinate reading will give a wrong focal length.

  • Forgetting to take the square root after obtaining f2=400f^2 = 400. The value 400400 is f2f^2, not ff. Therefore, compute f=400=20 cmf = \sqrt{400} = 20 \, \text{cm}.

Practice more Refraction & Lenses questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Refraction & Lenses

Related questions