An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

Engineering

Physics

SI Units

Plane Surface Refraction

Lens

Question

An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

3.2 m

5.6 m

7.2 m

2.4 m

$\frac{1}{12} + \frac{1}{240} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{7}{80}$

$Δx due to slab = t (1 - \frac{1}{μ}) = \frac{1}{3} CM$

* New v should be $= 12 - \frac{1}{3} = \frac{35}{3} CM$

* $\frac{1}{u} = \frac{1}{v} \frac{1}{f} = \frac{3}{35} \frac{7}{80} = - \frac{1}{500}$

u = –5.6 m

This problem involves finding the new object distance when a glass plate is inserted between a lens and its image plane. We'll solve it step by step.

Step 1: Find the Focal Length of the Lens

First, use the lens formula with the initial positions. The lens formula is:

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

Given:

Object distance, $u = - 2.4 m$ (negative sign by convention)
Image distance, $v = + 0.12 m$ (positive for real image)

Plug into the formula:

\frac{1}{f} = \frac{1}{0.12} - \frac{1}{- 2.4} = \frac{1}{0.12} + \frac{1}{2.4} = \frac{25}{3} + \frac{5}{12} = \frac{100}{12} + \frac{5}{12} = \frac{105}{12} = \frac{35}{4}

So, the focal length is:

f = \frac{4}{35} m \approx 0.1143 m

Step 2: Effect of the Glass Plate

When a parallel-sided glass slab is introduced in the path of a converging beam, it shifts the image away from the lens by a distance:

S h i f t = t (1 - \frac{1}{μ})

where:

$t = 1 c m = 0.01 m$ (thickness)
$μ = 1.50$ (refractive index)

Calculate the shift:

S = 0.01 (1 - \frac{1}{1.5}) = 0.01 (1 - \frac{2}{3}) = 0.01 \times \frac{1}{3} = \frac{0.01}{3} = \frac{1}{300} m

This shift is away from the lens, so the effective image distance for the lens (the position where the image would form without the plate to be sharp on the film) must now be:

v_{n e w} = v - S = 0.12 - \frac{1}{300} = \frac{36}{300} - \frac{1}{300} = \frac{35}{300} = \frac{7}{60} m

(The shift is subtracted because the film's position is fixed; to focus on it, the lens must form an image at a point that, after being shifted by the plate, lands on the film.)

Step 3: Find the New Object Distance

Now, use the lens formula again. We know $f$ and the new required image distance $v_{n e w}$ . Solve for the new object distance $u_{n e w}$ :

\frac{1}{f} = \frac{1}{v_{n e w}} - \frac{1}{u_{n e w}}

Substitute the known values ( $f = \frac{4}{35}$ , $v_{n e w} = \frac{7}{60}$ ):

\frac{1}{u_{n e w}} = \frac{1}{v_{n e w}} - \frac{1}{f} = \frac{1}{7 / 60} - \frac{1}{4 / 35} = \frac{60}{7} - \frac{35}{4} = \frac{240 - 245}{28} = \frac{- 5}{28}

Therefore, the new object distance is:

u_{n e w} = - \frac{28}{5} = - 5.6 m

The negative sign indicates it is an object distance. So, the magnitude is 5.6 m.

Final Answer

The object should be shifted to 5.6 m from the lens.

Detailed Solution

Step 1: Find the Focal Length of the Lens

Step 2: Effect of the Glass Plate

Step 3: Find the New Object Distance

Final Answer

Related Topics & Formulae

Detailed Solution

Step 1: Find the Focal Length of the Lens

Step 2: Effect of the Glass Plate

Step 3: Find the New Object Distance

Final Answer

Related Topics & Formulae

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