An object is placed at 21 cm in front of a concave mirror of radius of a curvature 10 cm. A glass slab of thickness 3 cm and u 15 is then placed close to the mirror in the space between the object and the mirror. The position of final image formed is

Options

A. `-3.94cm`
B. 4.3 cm
C. `-4.93`
D. 3.94 cm

Explanation

To solve the problem step-by-step, we will follow the outlined approach to find the position of the final image formed by the concave mirror after placing the glass slab.

### Step 1: Determine the focal length of the concave mirror
The radius of curvature (R) of the concave mirror is given as 10 cm. The focal length (f) is related to the radius of curvature by the formula:

\[ f = \frac{R}{2} \]

Substituting the value of R:

\[ f = \frac{10 \, \text{cm}}{2} = 5 \, \text{cm} \]

### Step 2: Calculate the increase in path due to the glass slab
The thickness (T) of the glass slab is given as 3 cm, and the refractive index (μ) is 1.5. The increase in path length (Δ) due to the glass slab is calculated using the formula:

\[ \Delta = (μ - 1) \cdot T \]

Substituting the values:

\[ \Delta = (1.5 - 1) \cdot 3 \, \text{cm} = 0.5 \cdot 3 \, \text{cm} = 1.5 \, \text{cm} \]

### Step 3: Calculate the new object distance (u')
The object is initially placed at 21 cm in front of the mirror. However, due to the glass slab, the effective object distance (u') will be modified by the increase in path length:

\[ u' = - (21 \, \text{cm} + 1.5 \, \text{cm}) = -22.5 \, \text{cm} \]

### Step 4: Apply the mirror formula to find the image distance (v)
The mirror formula is given by:

\[ \frac{1}{f} = \frac{1}{u'} + \frac{1}{v} \]

Rearranging the formula to find v:

\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u'} \]

Substituting the known values (f = 5 cm and u' = -22.5 cm):

\[ \frac{1}{v} = \frac{1}{5} - \frac{1}{-22.5} \]

Calculating the right side:

\[ \frac{1}{v} = \frac{1}{5} + \frac{1}{22.5} \]

Finding a common denominator (which is 22.5 * 5 = 112.5):

\[ \frac{1}{v} = \frac{22.5 + 5}{112.5} = \frac{27.5}{112.5} \]

Calculating v:

\[ v = \frac{112.5}{27.5} \approx 4.09 \, \text{cm} \]

### Step 5: Calculate the geometrical distance
Since the image distance is positive, it indicates that the image is formed on the same side as the object. However, we need to account for the thickness of the glass slab when determining the final geometrical distance from the mirror:

\[ \text{Geometrical distance} = v - \Delta = 4.09 \, \text{cm} - 1.5 \, \text{cm} = 2.59 \, \text{cm} \]

### Final Answer
The position of the final image formed is approximately **2.59 cm** in front of the mirror.

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