1. A thumb pin is positioned at a distance of $15\text{ cm}$ from a convex mirror of a focal length of $20\text{ cm}$. Determine the position and nature of the image.
Data:
- Object distance ($p$) = $+15\text{ cm}$
- Focal length ($f$) = $-20\text{ cm}$ (Negative for a convex mirror)
- Image distance ($q$) = ?
Solution:
Using the mirror formula:
$$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$$
$$\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$$
$$\frac{1}{q} = \frac{1}{-20} - \frac{1}{15} = -\frac{1}{20} - \frac{1}{15}$$
$$\frac{1}{q} = \frac{-3 - 4}{60} = \frac{-7}{60}$$
$$q = -\frac{60}{7} \approx -8.57\text{ cm}$$
RESULT: The image distance is $8.57\text{ cm}$ behind the mirror. The negative sign confirms the nature of the image is virtual and erect.
2. An image of a specimen appears to be $11.5\text{ cm}$ behind a concave mirror with a focal length of $13.5\text{ cm}$. Find the specimen's distance from the mirror.
Data:
- Image distance ($q$) = $-11.5\text{ cm}$ (Negative because the image is virtual / behind the mirror)
- Focal length ($f$) = $+13.5\text{ cm}$ (Positive for a concave mirror)
- Object distance ($p$) = ?
Solution:
Using the mirror formula:
$$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$$
$$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$$
$$\frac{1}{p} = \frac{1}{13.5} - \frac{1}{-11.5} = \frac{1}{13.5} + \frac{1}{11.5}$$
$$\frac{1}{p} \approx 0.07407 + 0.08696 = 0.16103$$
$$p = \frac{1}{0.16103} \approx 6.21\text{ cm}$$
RESULT: The specimen object distance from the mirror is $6.21\text{ cm}$.
3. A convex mirror used for rear-view on an automobile has a radius of curvature of $4.00\text{ m}$. If a bus is located at $5.00\text{ m}$ from this mirror, find the image's position, nature, and size.
Data:
- Radius of curvature ($R$) = $-4.00\text{ m}$
- Focal length ($f$) = $\frac{R}{2} = \frac{-4.00}{2} = -2.00\text{ m}$
- Object distance ($p$) = $+5.00\text{ m}$
- Image distance ($q$) = ?
- Magnification / Size ratio ($M$) = ?
Solution:
Part 1: Position of the Image
$$\frac{1}{f} = \frac{1}{p} + \frac{1}{q} \implies \frac{1}{q} = \frac{1}{f} - \frac{1}{p}$$
$$\frac{1}{q} = \frac{1}{-2.00} - \frac{1}{5.00} = -0.50 - 0.20 = -0.70$$
$$q = -\frac{1}{0.70} \approx -1.43\text{ m}$$
Part 2: Size of the Image (Magnification)
$$M = -\frac{q}{p} = -\frac{-1.43}{5.00} \approx 0.286$$
RESULT: The image is positioned $1.43\text{ m}$ behind the mirror. The nature of the image is virtual and erect, and its size is diminished to $0.286$ times the actual size of the bus.
Lenses and Refraction
4. An object is placed $15\text{ cm}$ away from a converging lens of a focal length of $10\text{ cm}$. Determine the position, size, and nature of the image formed.
Data:
- Object distance ($p$) = $+15\text{ cm}$
- Focal length ($f$) = $+10\text{ cm}$ (Positive for a converging/convex lens)
- Image distance ($q$) = ?
- Magnification ($M$) = ?
Solution:
Using the thin lens formula:
$$\frac{1}{f} = \frac{1}{p} + \frac{1}{q} \implies \frac{1}{q} = \frac{1}{f} - \frac{1}{p}$$
$$\frac{1}{q} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$$
$$q = 30\text{ cm}$$
Finding the size scalar via magnification:
$$M = \left| \frac{q}{p} \right| = \frac{30}{15} = 2$$
RESULT: The image is formed at a distance of $30\text{ cm}$ on the other side of the lens. The positive sign indicates it is a real and inverted image, magnified to twice ($2\times$) the original object size.
5. A concave lens of focal length $20\text{ cm}$ forms an image $15\text{ cm}$ from the lens. Determine the power of the lens. Also, how far is the object positioned from the lens?
Data:
- Focal length ($f$) = $-20\text{ cm} = -0.2\text{ m}$ (Negative for a diverging/concave lens)
- Image distance ($q$) = $-15\text{ cm} = -0.15\text{ m}$ (Diverging lenses always form virtual images)
- Power of the lens ($P$) = ?
- Object distance ($p$) = ?
Solution:
Part 1: Power of the Lens
$$P = \frac{1}{f\text{ (in meters)}} = \frac{1}{-0.2\text{ m}} = -5\text{ Diopters (D)}$$
Part 2: Object Distance
$$\frac{1}{f} = \frac{1}{p} + \frac{1}{q} \implies \frac{1}{p} = \frac{1}{f} - \frac{1}{q}$$
$$\frac{1}{p} = \frac{1}{-20} - \frac{1}{-15} = -\frac{1}{20} + \frac{1}{15}$$
$$\frac{1}{p} = \frac{-3 + 4}{60} = \frac{1}{60}$$
$$p = 60\text{ cm} = 0.6\text{ m}$$
RESULT: The power of the lens is $-5\text{ D}$, and the object is positioned $60\text{ cm}$ (or $0.6\text{ meters}$) in front of the lens.
6. The angle of incidence for a ray of light from air to water interface is $40^\circ$. If the ray travels through the water with a refractive index of $1.33$, calculate the angle of refraction.
Data:
- Angle of incidence ($i$) = $40^\circ$
- Refractive index of water ($n$) = $1.33$
- Angle of refraction ($r$) = ?
Solution:
According to Snell's Law for light entering from air ($n_{\text{air}} \approx 1$):
$$n = \frac{\sin(i)}{\sin(r)} \implies \sin(r) = \frac{\sin(i)}{n}$$
$$\sin(r) = \frac{\sin(40^\circ)}{1.33} \approx \frac{0.6428}{1.33} \approx 0.4833$$
$$r = \sin^{-1}(0.4833) \approx 28.9^\circ$$
RESULT: The angle of refraction inside the water medium is $28.9^\circ$.
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