One face of a rectangular glass plate 6 cm thick issilvered. An object...
Thickness of glass plate (t) = 6 cm;
Distance of the object (u) = 8 cm.
And distance of the image (v) = 12 cm.
Let x = Apparent position of the silvered
surface in cm.
Since the image is formed due to relfection at
the silvered face and by the property of mirror
image
Distance of object from the mirror = Distance
of image from the mirror
or, x + 8 = 12 + 6 – x ⇒ x = 5 cm.
Therefore, refractive index of glass
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One face of a rectangular glass plate 6 cm thick issilvered. An object...
The given problem involves a rectangular glass plate that is 6 cm thick and has one face that is silvered. An object is placed 8 cm in front of the silvered face, and the problem asks us to determine the refractive index of the glass.
To solve this problem, we will use the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
where:
v = image distance
u = object distance
f = focal length
We are given that the object is held 8 cm in front of the first face, so u = -8 cm (negative sign indicates that the object is on the same side as the incident light). We are also given that the image is formed 12 cm behind the silvered face, so v = -12 cm.
Now, let's consider the thickness of the glass plate. The light first enters the glass plate, then reflects off the silvered face, and finally exits the glass plate. This means that the light undergoes refraction twice, once when entering and once when exiting the glass.
Let's assume the refractive index of the glass is n. The refractive index of air is approximately 1. Therefore, when light enters the glass, it bends towards the normal. The angle of incidence is equal to the angle of refraction, so we can write:
$\sin i = n \sin r$
where i is the angle of incidence and r is the angle of refraction.
Similarly, when the light exits the glass, it bends away from the normal. Again, the angle of incidence is equal to the angle of refraction, so we can write:
$\sin i' = \frac{1}{n} \sin r'$
where i' is the angle of incidence and r' is the angle of refraction.
Since the angles of incidence and refraction are small, we can use the small angle approximation:
$\sin \theta \approx \theta$
Using this approximation, we can rewrite the equations as:
$i \approx n r$
$i' \approx \frac{r'}{n}$
Now, let's consider the geometry of the problem. The thickness of the glass plate is given as 6 cm, and the light travels a distance of 12 cm within the glass. Since the refractive index is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, we can write:
$n = \frac{c}{v_g}$
where c is the speed of light in vacuum and $v_g$ is the speed of light in the glass.
The time taken by light to travel a distance d is given by:
$t = \frac{d}{v}$
The speed of light is the distance traveled per unit time, so we can write:
$v = \frac{d}{t}$
Substituting the values of d = 12 cm and t = 6 cm into the equation, we get:
$v_g = \frac{12 \ cm}{6 \ cm} = 2$
Substituting the value of $v_g$ into the equation for refractive index, we get:
$n = \frac{c}{2}$
The speed of light in vacuum is approximately $3 \times 10^8 \ m/s$, so we can write:
$n = \frac{3 \times 10^