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An object is placed infront of a slab of thickness 10cm and refractive index =1.5 at a distance of 28cm from it.The other face of the slab is silvered.Find the final position of the image from the silvered face?
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An object is placed infront of a slab of thickness 10cm and refractive...
1. Refraction from the front face AB

2. Reflection from the silvered face

3. Again refraction from the front face AB

1. 28*1.5= 42cm

Hence the image will be formed at 42 +10= 52cm

I2 will be formed 52cm behind the mirror. 

i.e., 52+10= 62cm

Therefore final image I3will be at 62/1.5= 41.33cm
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Most Upvoted Answer
An object is placed infront of a slab of thickness 10cm and refractive...

Given Data:
- Thickness of the slab (t) = 10 cm
- Refractive index of the slab (n) = 1.5
- Distance of object from the slab (u) = 28 cm

Final Position of Image:


  • Finding the Image Distance:



The image distance (v) can be calculated using the lens formula:
1/f = 1/v - 1/u
where f is the focal length of the lens.

To find the focal length (f), we can use the formula:
f = t / (n - 1)

Substitute the values:
f = 10 / (1.5 - 1) = 20 cm

Now we can find the image distance (v) using the lens formula:
1/20 = 1/v - 1/28
1/v = 1/20 + 1/28
1/v = (7 + 5) / 140
1/v = 12 / 140
v = 140 / 12
v ≈ 11.67 cm

Therefore, the final position of the image from the silvered face of the slab is approximately 11.67 cm.
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An object is placed infront of a slab of thickness 10cm and refractive index =1.5 at a distance of 28cm from it.The other face of the slab is silvered.Find the final position of the image from the silvered face?
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