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Physics Case study (4 Marks)

Optical Instruments · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 3 questions.

Questions

Case study (4 Marks)

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3 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
A child has near point at 10cm. What is the maximum angular magnification the child can have with a convex lens of focal length 10cm?
Answer
The child has D = 10cm and f = 10cm. The maximum angular magnification is obtained when the image is formed at near point.$\text{m}=1+\frac{\text{D}}{\text{f}}=1+\frac{10}{10}=1+1=2$
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Question 24 Marks
By mistake, an eye surgeon puts a concave lens in place of the lens in the eye after a cataract operation. Will the patient be able to see clearly any object placed at any distance?
Answer
The image formed by a concave lens is virtual and upright. It is smaller than the object and is formed between the object and the lens, irrespective of the position of the object. If, by mistake, an eye surgeon puts a concave lens in place of eye lens in a patient's eye, then the image will not be focused on the retina; this will lead to unclear vision.
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Question 34 Marks
A lady cannot see objects closer than 40cm from the left eye and closer than 100cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telscope from her reading glasses to look for her teammates.
  1. Which glass should she use as the eyepiece?
  2. What magnification can she get with relaxed eye?
Answer
The lady can not see objects closer than 40cm from the left eye and 100cm from the right eye.For the left glass lens,
v = -40cm,
u = -25cm
$\therefore \frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-40}-\frac{1}{-25}=\frac{1}{25}-\frac{1}{40}=\frac{3}{200}$
$\Rightarrow \text{f}=\frac{200}{3}\text{cm}$
For the right glass lens,
v = -100cm,
u = -25cm
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-100}-\frac{1}{-25}=\frac{1}{100}=\frac{3}{100}$
$\Rightarrow\text{f}=\frac{100}{3}\text{cm}$
  1. For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should
use the right lens $\big(\text{f}=\frac{100}{3}\text{cm}\big)$ as the eye piece lens.
  1. With relaxed eye, (normal adjustment)
$\text{f}_0=\frac{200}{3}\text{cm}$
$\text{f}_\text{e}=\frac{100}{3}\text{cm}$
magnification $=\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{\big(\frac{200}{3}\big)}{\big( \frac{100}{3}\big)}=2$
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