Advertisements
Advertisements
प्रश्न
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Advertisements
उत्तर
Focal length of the convex lens, f1 = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f2
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as:
`1/"f" = 1/"f"_1 + 1/"f"_2`
`1/"f"_2 = 1/"f" - 1/"f"_1`
= `1/45 - 1/30`
= `-1/90`
∴ f2 = −90 cm
Let the refractive index of the lens be μ1 and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is −R.
R can be obtained using the relation:
`1/"f"_1 = (μ_1 - 1)(1/"R" + 1/(-"R"))`
`1/30 = (1.5 - 1) (2/"R")`
R = `30/(0.5 xx 2)`
∴ R = 30 cm
Let μ2 be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane mirror = ∞
Radius of curvature of the liquid on the side of the lens, R = −30 cm
The value of μ2 can be calculated using the relation:
`1/"f"_2 = (μ_2 - 1)[1/(-"R") - 1/∞]`
`(-1)/90 = (μ_2 - 1)[1/(+30) - 0]`
`μ_2 - 1 = 1/3`
μ2 = `1/3 + 1`
μ2 = `4/3`
∴ μ2 = 1.33
Hence, the refractive index of the liquid is 1.33.
संबंधित प्रश्न
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
An equiconvex lens of focal length 'f' is cut into two identical plane convex lenses. How will the power of each part be related to the focal length of the original lens ?
Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without loosing the energy of the light. This increase the intensity. Describe how the converging lenses should be placed to do this.
Answer the following question.
An optical instrument uses a lens of 100 D for the objective lens and 50 D for its eyepiece. When the tube length is kept at 20 cm, the final image is formed at infinity.
(a) Identify the optical instrument.
(b) Calculate the magnification produced by the instrument.
Show that the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.
A transparent refracting medium bounded by two surfaces, of which at least one is spherical, is called a:
The point on the principal axis where rays parallel to the principal axis actually meet after refraction, or appear to diverge after refraction, is called the:
The distance between the optical centre and the principal focus is called the:
According to the new Cartesian sign convention for lenses, the focal length of a convex lens is:
According to the new Cartesian sign convention used for lenses, distances measured opposite to the direction of incident light are taken as:
A ray of light passing through the optical centre of a thin lens:
A ray parallel to the principal axis, after refraction through a convex lens, passes through:
A ray passing through the principal focus of a convex lens, after refraction, emerges:
Using the sign convention with \[OB = -u\] and \[DI = +v\] in the combined refraction equation for a thin lens gives:
If the refractive index of a liquid equals that of a glass lens (\[n_1 = n_2\]), the lens maker relation gives \[\frac{1}{f} = 0\]. This means the focal length of the lens becomes:
A lens with an infinite focal length behaves like:
