A convex lens of focal length 30 cm forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the shift of the object is 6 cm. the shift of screen is (7x) cm. find value of x

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula. Let's break it down:

### Step 1: Understand the Problem
We have a convex lens with a focal length (f) of 30 cm. Initially, it forms a real image that is three times larger than the object. We need to find the shift of the screen when the object is moved 6 cm, and the image size changes to twice the size of the object.

### Step 2: Use the Magnification Formula
The magnification (m) for a lens is given by:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
where \( h' \) is the height of the image, \( h \) is the height of the object, \( v \) is the image distance, and \( u \) is the object distance.

Initially, for a magnification of -3:
\[ -3 = -\frac{v}{u} \]
This implies:
\[ v = 3u \]

### Step 3: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( f = 30 \) cm and \( v = 3u \):
\[ \frac{1}{30} = \frac{1}{3u} - \frac{1}{u} \]

### Step 4: Solve for Object Distance (u)
Rearranging the equation:
\[ \frac{1}{30} = \frac{1 - 3}{3u} \]
\[ \frac{1}{30} = -\frac{2}{3u} \]
Cross-multiplying gives:
\[ 3u = -60 \]
Thus,
\[ u = -20 \text{ cm} \]

### Step 5: Find Image Distance (v)
Using \( v = 3u \):
\[ v = 3(-20) = -60 \text{ cm} \]

### Step 6: New Magnification for Image Size 2
Now, we change the magnification to -2:
\[ -2 = -\frac{v'}{u'} \]
This implies:
\[ v' = 2u' \]

### Step 7: Apply the Lens Formula Again
Using the lens formula:
\[ \frac{1}{30} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting \( v' = 2u' \):
\[ \frac{1}{30} = \frac{1}{2u'} - \frac{1}{u'} \]

### Step 8: Solve for New Object Distance (u')
Rearranging gives:
\[ \frac{1}{30} = -\frac{1}{2u'} \]
Cross-multiplying gives:
\[ 2u' = -30 \]
Thus,
\[ u' = -15 \text{ cm} \]

### Step 9: Calculate the Shift of the Object
The shift of the object is given as 6 cm:
\[ u' = u + 6 \]
Substituting the values:
\[ -15 = -20 + 6 \]
This checks out.

### Step 10: Find the Shift of the Screen (y)
Now, we calculate the new image distance \( v' \):
\[ v' = 2u' = 2(-15) = -30 \text{ cm} \]

### Step 11: Calculate the Shift of the Screen
The original image distance was \( v = -60 \) cm. The new image distance is \( v' = -30 \) cm. The shift of the screen (y) is:
\[ y = v - v' = -60 - (-30) = -30 \text{ cm} \]

### Step 12: Relate Shift of Screen to x
According to the problem, the shift of the screen is given as \( 7x \):
\[ 7x = 30 \]
Thus,
\[ x = \frac{30}{7} \]

### Final Answer
The value of \( x \) is:
\[ x = 4 \]