A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Home » A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Class 10 | ICSE | Physics | Refraction of Light at Plane Surfaces | Selina

A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Solution:

According to the question, refractive index of the water (μ_w) is 4 / 3 and the real depth of coin is 12 cm

We have to find the shift in the image

Expression for the shift in image in terms of refractive index is given below –

Shift = Real depth × (1 – 1 / μ)

Putting values, Shift = 12 × (1 – 3 / 4)

Shift = 12 / 4

Therefore, Shift = 3 cm

i

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

Two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ , internal resistance $r_{1}$ and $r_{2}$ , connected in parallel. The equivalent emf of the combination is- (A) $\frac{\varepsilon_{1} r_{1}+\varepsilon_{1} r_{1}}{r_{1}+r_{2}}$ (B) $\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$ (C) $\sqrt{\varepsilon_{1} \times \varepsilon_{2}}$ (D) $\frac{\varepsilon_{1}+\varepsilon_{2}}{2}$

More Material

A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia.
A Concave, concave
B Convex, convex
C Convex, concave
D Concave, convex

Rate of reaction of any substance depends on
(A) active mass
(B) molecular weight
(C) atomic weight
(D) equivalent weight

Formula of oleum is
(A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$
(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$
(C) $\mathrm{H}_{2} \mathrm{SO}_{3}$
(D) $\mathrm{H}_{2} \mathrm{SO}_{5}$

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

If the points $A(a, 0), B(0, b)$ and $C(1,1)$ are collinear, prove that $\frac{1}{a}+\frac{1}{b}=1$

If $A(-2,0), B(0,4)$ and $C(0, k)$ be three points such that area of a $A B C$ is 4 sq units, find the value of $k$ .

Find the value of $k$ for which the area of a ABC having vertices $A(2,-6), B(5,4)$ and $C(k, 4)$ is 35 sq units.

Find the value of $k$ for which the points $A(1,-1), B(2, k)$ and $C(4,5)$ are collinear.

Find the value of $k$ for which the points $P(5,5), Q(k, 1)$ and $R(11,7)$ are collinear.

Find the value of $k$ for which the points $A(3,-2), B(k, 2)$ and $C(8,8)$ are collinear.

Use determinants to show that the following points are collinear. $P(-2,5), Q(-6,-7)$ and $R(-5,-4)$

Sign up now and study all subjects for free

Class

Syllabus

Question Papers

Social Media

Apps