Two thin similar watch glass pieces are joined together,front to front ,with rear convex portion silvered and the combination of glass pieces is placed at a distance `a=60cm` from a screen.A small object is placed normal to the optical axis of the combination such that its image is two time the object. Now water is filled in between the glass. water`mu=4//3`,calculated the distance through,which the object must be displayed so that a sharp image is again formed on the screen.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understanding the Setup
We have two thin watch glass pieces joined front to front, with the rear convex portion silvered, forming a concave mirror. The setup is initially filled with air, and an object is placed such that its image is twice the size of the object at a distance of 60 cm from the screen.

### Step 2: Analyzing the Initial Conditions
Given that the image is twice the size of the object, we can use the magnification formula:
\[ M = \frac{V}{U} \]
Where \(M = 2\). Thus,
\[ 2 = \frac{V}{U} \implies U = \frac{V}{2} \]
Since the image is formed at a distance of 60 cm from the mirror, we take \(V = -60\) cm (as per sign convention for mirrors). Therefore,
\[ U = \frac{-60}{2} = -30 \text{ cm} \]

### Step 3: Finding the Focal Length of the Concave Mirror
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{-60} + \frac{1}{-30} \]
Calculating:
\[ \frac{1}{f} = -\frac{1}{60} - \frac{2}{60} = -\frac{3}{60} = -\frac{1}{20} \]
Thus, the focal length \(f = -20\) cm. The radius of curvature \(R\) is:
\[ R = 2f = 2 \times (-20) = -40 \text{ cm} \]

### Step 4: Filling the Space with Water
Now, we fill the space with water, which has a refractive index \( \mu = \frac{4}{3} \). The lens maker's formula for a convex lens is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
Since both surfaces are convex and have the same radius of curvature \(R = 40\) cm, we have:
\[ \frac{1}{f} = \left(\frac{4}{3} - 1\right) \left( \frac{1}{40} + \frac{1}{40} \right) = \frac{1}{3} \times \frac{2}{40} = \frac{1}{60} \]
Thus, the focal length of the lens is \(f = 60\) cm.

### Step 5: Finding the New Object Distance
To find the new object distance \(u\) such that a sharp image is formed again at the screen (60 cm away), we need to consider the lens and mirror combination.

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where \(f = 60\) cm. The image formed by the lens will act as the object for the concave mirror.

Let \(v\) be the image distance from the lens. The image distance \(v\) will also satisfy the mirror formula:
\[ \frac{1}{f_m} = \frac{1}{v_m} + \frac{1}{u_m} \]
Where \(f_m = -20\) cm and \(v_m = -60\) cm (the screen position).

### Step 6: Solving for Object Distance
Substituting into the lens formula:
\[ \frac{1}{60} = \frac{1}{v} - \frac{1}{u} \]
And for the mirror:
\[ \frac{1}{-20} = \frac{1}{-60} + \frac{1}{u'} \]
Where \(u'\) is the object distance for the mirror.

From the lens formula, we can express \(v\):
\[ v = \frac{60u}{60 + u} \]

Substituting \(v\) into the mirror equation:
\[ \frac{1}{-20} = \frac{1}{-60} + \frac{60 + u}{60u} \]
Solving this gives:
\[ -3 + \frac{1}{60} = \frac{60 + u}{60u} \]
This simplifies to:
\[ -1/30 = \frac{60 + u}{60u} \]
Cross multiplying and solving for \(u\):
\[ -2u = 60 + u \implies 3u = -60 \implies u = -20 \text{ cm} \]

### Step 7: Displacement Calculation
The object was initially at \(U = -30\) cm and must now be at \(U = -20\) cm. Thus, the object must be displaced by:
\[ 30 - 20 = 10 \text{ cm} \]

### Final Answer
The object must be displaced by **10 cm**.

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