Light and Optics

Free sample questions

Light and Optics — sample Physics questions with answers

1. Question 1

   A swimming pool has an actual depth of $2.4\text{ m}$. If the refractive index of water is $1.33$, what is the apparent depth of the pool when viewed vertically from above?

   • A$3.19\text{ m}$
   • B$1.80\text{ m}$
   • C$1.07\text{ m}$
   • D$0.55\text{ m}$

   Answer: B. $1.80\text{ m}$

   Using the refractive index formula, $n=\frac{\text{Actual depth, }D}{\text{Apparent depth, }d}$. Hence, $d=\frac{D}{n}=\frac{2.4\text{ m}}{1.33}\approx 1.80\text{ m}$.

2. Question 2

   The refractive index of a glass block is $1.52$. Calculate the critical angle, $c$, of the glass.

   • A$48.75^{\circ }$
   • B$30.00^{\circ }$
   • C$41.14^{\circ }$
   • D$58.86^{\circ }$

   Answer: C. $41.14^{\circ}$

   The critical angle, $c$, is calculated using the formula $\sin c=\frac{1}{n}$. Therefore, $\sin c=\frac{1}{1.52}\approx 0.6579$, which gives $c={\sin }^{-1}(0.6579)\approx 41.14^{\circ }$.

3. Question 3

   An object is placed at a distance of $15\text{ cm}$ in front of a convex lens with a focal length of $10\text{ cm}$. What is the image distance, $v$, formed and what is the nature of the image?

   • A$v=6\text{ cm}$, real
   • B$v=30\text{ cm}$, virtual
   • C$v=6\text{ cm}$, virtual
   • D$v=30\text{ cm}$, real

   Answer: D. $v = 30\text{ cm}$, real

   Using the thin lens formula: $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$. Substitute $f=+10\text{ cm}$ and $u=+15\text{ cm}$: $\frac{1}{10}=\frac{1}{15}+\frac{1}{v}⟹\frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}$. Thus, $v=30\text{ cm}$. Since the value of $v$ is positive, the image formed is real.

4. Question 4

   A light ray travels from air into a glass block with refractive index $1.50$. If the angle of incidence is $30^{\circ }$, what is the angle of refraction in the glass?

   • A$19.5^{\circ }$
   • B$20.0^{\circ }$
   • C$45.0^{\circ }$
   • D$48.6^{\circ }$

   Answer: A. $19.5^{\circ}$

   Using Snell's law: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$. Thus, $1.00\sin 30^{\circ }=1.50\sin {\theta }_2$, so $\sin {\theta }_2=0.333$, giving ${\theta }_2\approx 19.5^{\circ }$.

5. Question 5

   The phenomenon of total internal reflection occurs when light travels from a denser medium to a less dense medium. Which condition must be satisfied for this phenomenon to occur?

   • AThe angle of incidence is smaller than the critical angle
   • BThe angle of incidence is greater than the critical angle
   • CThe refractive index of the second medium is higher
   • DLight travels along the normal

   Answer: B. The angle of incidence is greater than the critical angle

   Total internal reflection occurs when light travels from a denser to a less dense medium and the angle of incidence exceeds the critical angle, $c$.