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Chapter 18: Problem 23

An object is placed at a distance of \(3 f\) from a convex lens of focal length \(f\). A slab of refractive index \(\mu\) is placed in between lens and object. The image of the object will be formed nearest to the object if thickness of the slab is (A) \(f\) (B) \(2 f\) (C) \(\frac{f}{\mu-1}\) (D) \(\frac{\mu f}{\mu-1}\)

Short Answer

Expert verified
The thickness of the slab that would cause the image to form nearest to the object is given by (C) \[\frac{f}{\mu-1}\].

Step by step solution

01

List given information

The object distance from the lens (u) = -3f (negative sign because it is a real object) The focal length of the lens (f) = f The refractive index of the slab (渭) = 渭
02

Write the lens formula

The lens formula is given as \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\] Where, f = focal length u = object distance from the lens v = image distance from the lens (this is what we need to find)
03

Write the refractive index formula and relation with thickness

Refraction of light occurs as the light passes through the slab. The refractive index formula is given as: \[\mu = \frac{d}{t}\] Where, 渭 = refractive index of the slab d = distance the image appears to have moved due to refraction t = thickness of the slab Since the lens and the slab are in contact, we can establish a relationship between d and t: \[d = t(\mu - 1)\]
04

Update the object distance based on refraction

Due to the refraction caused by the slab, the image appears to form at a shifted position. Let's call the shifted object distance u': \[u' = u + d\] By substituting the formula for d from Step 3, we get: \[u' = u + t(\mu - 1)\]
05

Find the image distance nearest to the object

Using the lens formula with u' as the object distance, we can solve for the image distance v. \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u'}\] \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u + t(\mu - 1)}\] Rearrange the formula to solve for v: \[v = \frac{1}{\frac{1}{f} + \frac{1}{u + t(\mu - 1)}}\]
06

Minimize image distance v with respect to thickness t

Now, we want to find the thickness t that minimizes the image distance v. To do this, we take the derivative of v with respect to t and set it to zero. \[\frac{dv}{dt} = 0\]
07

Solve for the thickness t

Solve the equation from Step 6 for t: \[t = \frac{f}{\mu - 1}\] Comparing this result with the options given in the exercise, we see that the correct answer is: (C) \[\frac{f}{\mu-1}\]

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Most popular questions from this chapter

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