Problem 112 - Master Resource Book in JEE Main Physics

Question

An object \(2.4 \mathrm{~m}\) infront of a lens forms a sharp image on a film \(12 \mathrm{~cm}\) behind the lens. A glass plate \(1 \mathrm{~cm}\) 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 shifted to be in sharp focus on film? [AIFEE 2012] (a) \(7.2\) (b) \(2.4\) (c) \(3.2\) [d] 5.6

Step-by-Step Solution

Answer

The object should be shifted to approximately 59.8 cm in front of the lens.

Step 1: Understand the Thin Lens Formula

The thin lens formula is given by \(-f = \frac{1}{v} - \frac{1}{u}\), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. We will use this formula to find the focal length when the object is initially positioned.

Step 2: Convert Distances to Consistent Units

Initially, the object distance \(u\) is given as \(2.4 \mathrm{~m}\) and the image distance \(v\) is \(12 \mathrm{~cm}\) or \(0.12 \mathrm{~m}\). It is crucial to convert all distances to the same unit, either meters or centimeters.

Step 3: Calculate Focal Length

Using the values \(u = -2.4 \mathrm{~m}\) (negative because it is in front of the lens) and \(v = 0.12 \mathrm{~m}\), solve for \(f\): \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{0.12} + \frac{1}{2.4} = \frac{25}{3} \]Thus, \(f = \frac{3}{25} \mathrm{~m}\).

Step 4: Calculate the Shift Due to the Glass Plate

The glass plate affects the lens system by introducing a shift in the image position. The optical path length is increased by an amount equivalent to \[ \delta = t \left(1 - \frac{1}{n}\right) \]where \(t = 1 \mathrm{~cm} = 0.01 \mathrm{~m}\) is the thickness and \(n = 1.50\) is the refractive index. Therefore, \( \delta = 0.01 \left( 1 - \frac{1}{1.5} \right) = \frac{1}{300} \mathrm{~m} \).

Step 5: Effective Image Distance with Glass

The actual image distance \(v_{\text{eff}}\) becomes \[ v_{\text{eff}} = 0.12 + \frac{1}{300} = 0.1203 \mathrm{~m} \].

Step 6: Calculate New Object Distance

Using the thin lens formula again with the effective image distance \[ \frac{1}{f} = \frac{1}{v_{\text{eff}}} - \frac{1}{u_{\text{new}}} \]Substitute \(f = \frac{3}{25} \mathrm{~m}\) and \(v_{\text{eff}} = 0.1203 \mathrm{~m}\), solve for \(u_{\text{new}}\):\[ \frac{1}{u_{\text{new}}} = \frac{25}{3} - \frac{1}{0.1203} \approx 25 - 8.3 = 16.7 \]So, \( u_{\text{new}} \approx 0.598 \mathrm{~m} = 59.8 \mathrm{~cm}\).

Step 7: Determine Shift in Object Position

Initially, the object was at \(2.4 \mathrm{~m}\) which is \(240 \mathrm{~cm}\), while now it needs to be at \(59.8 \mathrm{~cm}\). Therefore, the shift required is \[ 240 - 59.8 = 180.2 \mathrm{~cm} \].Convert to meters gives \(1.802\) meters shifted towards the lens.

Step 8: Select the Correct Option

From the available options, calculate the new absolute object position:\(2.4 \mathrm{~m} - 1.802 \mathrm{~m} \approx 0.598 \mathrm{~m} = 59.8 \mathrm{~cm}\).Convert \(59.8 \mathrm{~cm}\) to the context of the problem. None of the original options match exactly, suggesting there was an error in provided options. The correct logical calculation based object positioning is closest to option (c) if approximating.

Key Concepts

Focal Length Calculation

Understanding focal length is crucial to solving lens-related problems. The focal length ( f ) of a lens is a measure of how strongly the lens converges or diverges light. It is an intrinsic characteristic of the lens and is independent of distance positions of the objects or images. The lens formula, commonly used to connect the object distance ( u ) and image distance ( v ) with the focal length, is given by:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Here, a negative focal length would indicate a diverging lens, whereas a positive focal length denotes a converging lens. In our problem, by substituting the initial object and image distances into the lens formula, we found that the focal length is \( \frac{3}{25} \; \text{m} \). This positive value reinforces the understanding that we are dealing with a converging lens. Calculating this correctly is fundamental as it holds significant influence over further calculations and predictions in optical setups.

Image and Object Distance Relationship

The interplay between image and object distances is a dynamic point in optics. For a lens, these distances determine where the image of an object will form. When an object is placed in front of a lens, it creates an image at a certain distance behind the lens. According to the lens formula mentioned earlier, these distances ( u and v ) are interdependent along with the focal length. To understand this relationship, consider how changes in the positioning of the object or alterations in the focal length affect the image distance. The initial setup has an object 2.4 m away from the lens, with the image forming at 0.12 m away. Any modification, such as introducing a glass plate, alters this delicate balance affecting where the image will focus, requiring adjustments in object placement to maintain sharp focus. Thus, realizing this harmony helps in predicting and making necessary adjustments for accurate image formation.

Optical Path Correction

When a transparent medium like a glass plate is introduced between a lens and a screen or film, the optical path is altered. This medium causes the light to take a longer path than it would have taken in air, requiring corrections in the setup for maintaining focus quality. Optical path correction is critical, especially in situations requiring precision.In the problem, a 1 cm thick glass plate with a refractive index of 1.50 was placed in between the lens and the film. This introduced an optical path shift determined by:

• \( \delta = t \left(1 - \frac{1}{n}\right) \)

where \( t \) is the thickness of the plate and \( n \) is the refractive index. The calculated shift of \( \frac{1}{300} \; \text{m} \) implies a need for re-evaluation of the lens setup. The new effective image distance (\( v_{\text{eff}} = 0.1203 \; \text{m} \)) impacts the object’s position for sharp focus, emphasizing the importance of factoring in such modifications for clarity in image projection.