A convex lens focuses an object 40cm from it on a screen placed 10cm away from it. A glass plate (`mu=1.5`) and of thickness 3cm is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen ?

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Understand the initial setup
We have a convex lens that focuses an object placed 40 cm away from it onto a screen that is 10 cm away from the lens.

### Step 2: Identify the given values
- Object distance (u) = -40 cm (the negative sign indicates that the object is on the opposite side of the light direction)
- Image distance (v) = 10 cm (the positive sign indicates that the image is on the opposite side of the lens)

### Step 3: Use the lens formula to find the focal length (f)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{10} - \frac{1}{-40} \]
Calculating:
\[ \frac{1}{f} = \frac{1}{10} + \frac{1}{40} \]
To add these fractions, find a common denominator (40):
\[ \frac{1}{f} = \frac{4}{40} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8} \]
Thus, the focal length \( f = 8 \, \text{cm} \).

### Step 4: Calculate the shift caused by the glass plate
The glass plate has a refractive index \( \mu = 1.5 \) and a thickness of 3 cm. The shift (d) caused by the glass plate can be calculated using the formula:
\[ d = t \left(1 - \frac{1}{\mu}\right) \]
Substituting the values:
\[ d = 3 \left(1 - \frac{1}{1.5}\right) = 3 \left(1 - \frac{2}{3}\right) = 3 \left(\frac{1}{3}\right) = 1 \, \text{cm} \]

### Step 5: Adjust the image distance due to the shift
The original image distance was 10 cm. With the shift of 1 cm, the new image distance (v') becomes:
\[ v' = 10 \, \text{cm} - 1 \, \text{cm} = 9 \, \text{cm} \]

### Step 6: Use the lens formula again to find the new object distance (u')
Now we need to find the new object distance (u') such that the image is still focused on the screen. Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting the values:
\[ \frac{1}{8} = \frac{1}{9} - \frac{1}{u'} \]
Rearranging gives:
\[ \frac{1}{u'} = \frac{1}{9} - \frac{1}{8} \]
Finding a common denominator (72):
\[ \frac{1}{u'} = \frac{8}{72} - \frac{9}{72} = -\frac{1}{72} \]
Thus, \( u' = -72 \, \text{cm} \).

### Final Answer
The object should be placed at a distance of **72 cm** from the lens (on the same side as the original object).

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