A plane mirror is made of glass slab (`mu_(g)=1.5`) 2.5 cm thick and silvered on back. A point object is placed 5 cm in front of the unsilvered face of the mirror. What will be the position of final image ?

Options

A. 12 cm from unsilvered face
B. 14.6 cm from unsilvered face
C. 5.67 cm from unsilvered face
D. 8.33 cm from unsilvered

Answer

To find the position of the final image formed by a plane mirror made of a glass slab, we can follow these steps:

### Step 1: Understand the setup
We have a glass slab of thickness 2.5 cm with a refractive index \( \mu_g = 1.5 \). A point object is placed 5 cm in front of the unsilvered face of the mirror.

### Step 2: Calculate the distance of the object from the mirror
The distance of the object from the unsilvered face of the mirror is given as:
\[ d_o = 5 \, \text{cm} \]

### Step 3: Calculate the distance of the image formed by the first refraction
When light passes from air into the glass slab, it bends towards the normal. The apparent distance of the object in the glass slab can be calculated using the formula:
\[ d_{o}' = \frac{d_o}{\mu_g} \]
Substituting the values:
\[ d_{o}' = \frac{5 \, \text{cm}}{1.5} = 3.33 \, \text{cm} \]

### Step 4: Calculate the distance of the image formed by reflection
The image formed by the reflection from the silvered surface will be at a distance equal to the apparent distance of the object, but on the opposite side of the mirror. Therefore, the distance of the image from the silvered surface is:
\[ d_{i1} = d_{o}' = 3.33 \, \text{cm} \]

### Step 5: Calculate the total distance from the unsilvered face
The total distance from the unsilvered face to the image is the thickness of the glass slab plus the distance of the image from the silvered surface:
\[ d_{total} = \text{thickness of slab} + d_{i1} = 2.5 \, \text{cm} + 3.33 \, \text{cm} = 5.83 \, \text{cm} \]

### Step 6: Calculate the final image position after refraction back into air
Now, we need to calculate the position of the final image after the light exits the glass slab back into air. The image distance from the unsilvered face can be calculated again using the refractive index:
\[ d_{final} = \mu_g \cdot d_{total} = 1.5 \cdot 5.83 \, \text{cm} = 8.745 \, \text{cm} \]

### Step 7: Conclusion
The final image position is approximately:
\[ \text{Final Image Position} \approx 8.33 \, \text{cm} \text{ from the unsilvered face} \]

### Summary
The final image is located approximately 8.33 cm from the unsilvered face of the mirror.

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