Answer: Option D. -> f=−15 cm
:
D
Given:
Image distance $v=-10cm$
Object distance $u=-30cm$
(Using the sign conventions)
Let focal length of the lens be $f$.
Using the lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{f}=\left(\frac{1}{-10}\right)-\left(\frac{1}{-30}\right)$
$f=-15cm$
Here, negative sign for focal length shows that the lens is concave.

Answer: Option B. -> 30 m
:
B
Given:
Radius of curvature, $R=-40m$
Object distance, $u=-60m$
Let the sign convention be such that the object distance is negative i.e. positive direction is taken along the direction of the incident ray.
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
where:
$v:$ Position of image
$f:$ Focal length
But focal length is half of the radius.
$⟹f=\frac{R}{2}=-20m$
$⟹\frac{1}{v}+\frac{1}{-60}=\frac{1}{-20}$
$v=-30m$
Negative sign represents that object andimage are on the same side of mirror.

Answer: Option C. -> appear to come from F 
:
C

Ray parallel to principal axis after hitting the convex mirror will appear to come from its focus.

Answer: Option D. -> 25 ms−1
:
D
Given:
Speed of object, $u_o=5m{s}^{-1}$
Speed of mirror, $u_m=10m{s}^{-1}$
Mirror and object are approaching each other.
To find:
Speed of image
Procedure:
Let initial position of object be at $x_{oi}=0$
Let initial position of mirror be at $x_{mi}=u$
Object distance equals the image distance. Hence, initial position of image is $x_{ii}=2u$
Distance is the product of speed and time.
After time $t$,
Position of object will be:
$x_{of}=x_{oi}+u_{o}t=5t$
Position of mirror will be:
$x_{mf}=x_{mi}-u_{f}t=u-10t$
Let the final position of image be $x_{if}$
Using object distance equals image distance after time $t$.
$x_{if}-x_{mf}=x_{mf}-x_{of}$
$x_{if}=2u-25t$
Distance moved by image in time $t$ is:
$s_i=x_{if}-x_{ii}$
$s_i=-25t$
Speed is the ratio of distance and time. Speed of the image is:
$v_i=\frac{s_i}{t}=-25$
Negative sign appears because image is moving leftwards.
Hence, speed of image is $25m{s}^{-1}$

Answer: Option D. -> 15 cm
:
D
Given:
Refractive index of water, ${\mu }_{water}$ ​= $\frac{4}{3}$
Real depth of fish from surface of water$=20cm$
In the given situation,
Refractive Index of water $={\mu }_{water}=\frac{Realdepth}{Apparentdepth}$
$\Rightarrow Apparentdepth=\frac{Realdepth}{{\mu }_{water}}$
$\Rightarrow Apparentdepth=\frac{20}{\frac{4}{3}}$
Therefore, $Apparentdepth=\frac{20\times 3}{4}$ = 15 cm. The eagle will perceive the fish to be at $15cm$ depth from the water surface.

Answer: Option A. -> True
:
A
The rays of light travelling parallel to the principal axis, after reflection from a concave mirror meet at a single point onlyif the beam of light is narrow or if the mirror is of small aperture. In case, a wide beam of light falls on a concave mirror of large aperture, the rays after reflection would not converge at a single point. Therefore, if the aperture of the mirror is small, the image formed will be sharper.

Answer: Option B. -> 5 cm  
:
B
Given, Radius of curvature $=10cm$
If theobject is placed at infinity the image is formed at thefocus.
We know the focal length(F) is half the radius of curvature(R).
$F=\frac{R}{2}$
$F=\frac{10}{2}$
$⟹$ $F=5cm$
So the image of thesun is at 5 cm from the pole.

Answer: Option B. -> Increases
:
B
Magnification is given by:
$\text{Magnification}=\frac{\text{Height of image}}{\text{Height of object}}$
At center of curvature,
Image size equals the object size. Magnitude of magnification is unity.
At focus,
Image size equals infinity.Magnitude of magnification is infinity.
Therefore, it can be concluded that the magnification is increasing.

Answer: Option B. -> 2 cm
:
B
Given:
Object size, $h_o=4cm$
Object distance, $u=-30cm$
Imagedistance, $v=15cm$
Let image size be $h_i$.
Magnification of a lens is given by:
$m=\frac{v}{u}=\frac{h_i}{h_o}$
$\frac{+15}{-30}=\frac{h_i}{4}$
$h_i=-2cm$
Hence, image height is $2cm$ and negative sign shows that the image is inverted.

Answer: Option C. -> Concave lens,  $f=-2m$
:
C
Let $u$ be the object distance, $v$ be the image distance,
and $f$ be the focal length.
Given:
As per sign conventions, $u=-2m$ 
Magnification $m=\frac{1}{2}$    (Since size of image is half of that of object)
Magnification in lenses $m=\frac{v}{u}$
$\frac{1}{2}=\frac{v}{u}$
We get $v=\frac{u}{2}$
  $\Rightarrow v=\frac{-2}{2}=-1m$
Using lens formula ,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$\frac{1}{-1}-\frac{1}{-2}=\frac{1}{f}$

$\frac{1}{2}-1=\frac{1}{f}$

$f=-2m$
Here negative sign indicates that the lens is a concave lens.